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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 36 "An introduction to counting metho
ds " }}{PARA 0 "" 0 "" {TEXT -1 37 "by Peter Stone, Nanaimo, B.C., Can
ada" }}{PARA 0 "" 0 "" {TEXT -1 18 "Version: 23.3.2007" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "
" 0 "" {TEXT -1 28 "The multiplication principle" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 262 8 "Question" }
{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 209 "A particular model of car is available in a choice of fo
ur colours: gold, white, green, and maroon, and it can come with eithe
r manual or automatic transmission. How many different choices of car \+
are possible?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 263 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 111 "With each of the four colour choices
 we can choose either manual and automatic transmission making a total
 of  " }{XPPEDIT 18 0 "4*`.`* 2 = 8" "6#/*(\"\"%\"\"\"%\".GF&\"\"#F&\"
\")" }{TEXT -1 9 " choices." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
256 "" 0 "" {XPPEDIT 18 0 "matrix([[gold, manual], [gold, automatic], \+
[white, manual], [white, automatic], [green, manual], [green, automati
c], [maroon, manual], [maroon, automatic]]);" "6#-%'matrixG6#7*7$%%gol
dG%'manualG7$F(%*automaticG7$%&whiteGF)7$F-F+7$%&greenGF)7$F0F+7$%'mar
oonGF)7$F3F+" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 2 "A " }{TEXT 259 12 "tree diagram" }{TEXT -1 39 " \+
can be used to illustrate the answer. " }}{PARA 256 "" 0 "" {TEXT -1 
1 " " }{GLPLOT2D 383 370 370 {PLOTDATA 2 "6A-%'CURVESG6$7$7$$\"\"!F)F(
7$$\"\"\"F)F+-%&COLORG6&%$RGBGF(F(F(-F$6$7$F'7$F+$\"\"$F)F--F$6$7$F'7$
F+$!\"\"F)F--F$6$7$F'7$F+$!\"$F)F--F$6$7$F47$$\"\"#F)$\"3++++++++N!#<F
--F$6$7$F47$FG$\"3++++++++DFKF--F$6$7$F*7$FG$\"3++++++++:FKF--F$6$7$F*
7$FG$\"3++++++++]!#=F--F$6$7$F@7$FG$!3++++++++NFKF--F$6$7$F@7$FG$!3+++
+++++DFKF--F$6$7$F:7$FG$!3++++++++:FKF--F$6$7$F:7$FG$!3++++++++]FhnF--
F$6&7/F'F4F*F:F@FFFOFUFenF^pFhoFboF\\o-%'COLOURG6&F0$\"*++++\"!\")F(F(
-%'SYMBOLG6$%'CIRCLEG\"#5-%&STYLEG6#%&POINTG-F$6&FcpFdp-F[q6$%(DIAMOND
GF^qF_q-F$6&FcpFdp-F[q6$%&CROSSGF^qF_q-F$6&Fcp-Fep6&F0F)F)F)-F[q6$F]q
\"#7F_q-%%TEXTG6%7$$\"\")F<$\"#>F<Q%gold6\"-F.6&F0$F,!\"#F`sF`s-Fer6%7
$$\"\"(F<$\"\"&F<Q&whiteF]sF^s-Fer6%7$Fes$FBF<Q&greenF]sF^s-Fer6%7$$\"
\"*F<$FasF)Q'maroonF]sF^s-Fer6%7$$\"#:F<$\"#OF<Q'manualF]sF^s-Fer6%7$F
it$\"#DF<Q*automaticF]sF^s-Fer6%7$Fit$\"#;F<F]uF^s-Fer6%7$FitFgsFcuF^s
-Fer6%7$Fit$!\"%F<F]uF^s-Fer6%7$Fit$!#:F<FcuF^s-Fer6%7$Fit$!#CF<F]uF^s
-Fer6%7$Fit$!#NF<FcuF^s-%*AXESSTYLEG6#%%NONEG-%+AXESLABELSG6%Q!F]sFgw-
%%FONTG6#%(DEFAULTG-%%VIEWG6$F[xF[x" 1 2 0 1 10 0 2 9 1 1 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve
 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve 11" "Cur
ve 12" "Curve 13" "Curve 14" "Curve 15" "Curve 16" "Curve 17" "Curve 1
8" "Curve 19" "Curve 20" "Curve 21" "Curve 22" "Curve 23" "Curve 24" "
Curve 25" "Curve 26" "Curve 27" "Curve 28" }}{TEXT -1 1 " " }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 10 "The basic " }{TEXT 259 19 "multiplication rule" }
{TEXT -1 28 " for counting is as follows:" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT 258 29 "If one choice can be made in " }
{TEXT 271 1 "r" }{TEXT 260 5 " ways" }{TEXT 258 51 ", and a second cho
ice can be made independently in " }{TEXT 270 1 "s" }{TEXT 260 5 " way
s" }{TEXT 258 49 ", then the two choices can be made in a total of " }
{TEXT 272 1 "r" }{TEXT 260 3 " . " }{TEXT 273 1 "s" }{TEXT 260 5 " way
s" }{TEXT -1 1 "." }}{PARA 256 "" 0 "" {TEXT 264 28 "_________________
___________" }}{PARA 0 "" 0 "" {TEXT -1 23 "Try questions 1 and 2. " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 15 
"Arrangement of " }{TEXT 274 1 "n" }{TEXT -1 17 " things in a line" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 
265 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 88 "In ho
w many ways can Albert, Betsy, Chris and David line up to have their p
icture taken?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 266 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 59 "Let A, B, C and D represent Albert, B
etsy, Chris and David." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 15 "" 
0 "" {TEXT -1 8 "We have " }{TEXT 259 1 "4" }{TEXT -1 79 " choices for
 the person in the left-hand or 1st position, namely A, B, C and D." }
}{PARA 15 "" 0 "" {TEXT -1 59 "Once the1st position on the left has be
en filled there are " }{TEXT 259 1 "3" }{TEXT -1 214 " ways of choosin
g someone for the 2nd position to the right of the1st position. For ex
ample, if C occupies the 1st position on the left, then we could choos
e A, B or D for the second position. Altogether there are " }{XPPEDIT 
18 0 "4*`.`*3 = 12" "6#/*(\"\"%\"\"\"%\".GF&\"\"$F&\"#7" }{TEXT -1 41 
" ways of filling the first two positions." }}{PARA 15 "" 0 "" {TEXT 
-1 92 " Once the first two positions on the left are filled, we have t
wo people left, so there are " }{TEXT 256 1 "2" }{TEXT -1 55 " ways of
 filling the 3rd position. There are therefore " }{XPPEDIT 18 0 "12*`.
`*2 = 24" "6#/*(\"#7\"\"\"%\".GF&\"\"#F&\"#C" }{TEXT -1 43 " ways of f
illing the first three positions." }}{PARA 15 "" 0 "" {TEXT -1 112 "Th
e last position on the right must then be occupied by the remaining pe
rson, that is, it can be filled in only " }{TEXT 256 1 "1" }{TEXT -1 
5 " way." }}{PARA 15 "" 0 "" {TEXT -1 31 "The four people can line up \+
in " }{XPPEDIT 18 0 "4*`.`*3*`.`*2*`.`*1 = 24" "6#/*0\"\"%\"\"\"%\".GF
&\"\"$F&F'F&\"\"#F&F'F&F&F&\"#C" }{TEXT -1 6 " ways." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "The 24 possibilities ar
e:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 " A \+
B C D, A B D C, A C B D, A C D B, A D B C, A D C B," }}{PARA 0 "" 0 "
" {TEXT -1 54 " B A C D, B A D C, B C A D, B C D A, B D A C, B D C A,
" }}{PARA 0 "" 0 "" {TEXT -1 54 " C A B D, C A D B, C B A D, C B D A, \+
C D A B, C D B A," }}{PARA 0 "" 0 "" {TEXT -1 54 " D A B C, D A C B, D
 B A C, D B C A, D C A B, D C B A." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 36 "Similarly, 5 people can line up in  " }
{XPPEDIT 18 0 "5*`.`*4*`.`*3*`.`*2*`.`*1  = 120" "6#/*4\"\"&\"\"\"%\".
GF&\"\"%F&F'F&\"\"$F&F'F&\"\"#F&F'F&F&F&\"$?\"" }{TEXT -1 118 " ways b
ecause the positions from left to right can be filled in 5 ways, 4 way
s, 3 ways, 2 ways and 1 way respectively." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "It is convenient to use the " }
{TEXT 259 9 "factorial" }{TEXT -1 59 " notation to represent products \+
of the type occurring here." }}{PARA 0 "" 0 "" {TEXT -1 34 "If n is a \+
positive integer, then  " }{XPPEDIT 18 0 "n! = n*`. `*(n-1)*`.`*(n-2)*
` . . . `*3*`.`*2*`.`*1" "6#/-%*factorialG6#%\"nG*8F'\"\"\"%#.~GF),&F'
F)F)!\"\"F)%\".GF),&F'F)\"\"#F,F)%(~.~.~.~GF)\"\"$F)F-F)F/F)F-F)F)F)" 
}{TEXT -1 4 ". ( " }{TEXT 276 2 "n!" }{TEXT -1 12 " is read as " }
{TEXT 275 1 "n" }{TEXT 259 10 " factorial" }{TEXT -1 2 ")." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "1! = 1" "6#/-%*
factorialG6#\"\"\"F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 
"2! = 2*`.`*1" "6#/-%*factorialG6#\"\"#*(F'\"\"\"%\".GF)F)F)" }{TEXT 
-1 5 " = 2 " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "3! = 3*`.`*2*`.`* 1" "6#
/-%*factorialG6#\"\"$*,F'\"\"\"%\".GF)\"\"#F)F*F)F)F)" }{TEXT -1 5 " =
 6 " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "4! = 4*`.`*3*`.`*2*`.`*1" "6#/-%
*factorialG6#\"\"%*0F'\"\"\"%\".GF)\"\"$F)F*F)\"\"#F)F*F)F)F)" }{TEXT 
-1 6 " = 24 " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "5! = 5*`.`*4*`.`*3*`.`*
2*`.`*1;" "6#/-%*factorialG6#\"\"&*4F'\"\"\"%\".GF)\"\"%F)F*F)\"\"$F)F
*F)\"\"#F)F*F)F)F)" }{TEXT -1 7 " = 120 " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "for n from 1 to 15 do
\n   printf(\"%d! = %d\\n\",n,n!);\nend do:" }}{PARA 6 "" 1 "" {TEXT 
-1 6 "1! = 1" }}{PARA 6 "" 1 "" {TEXT -1 6 "2! = 2" }}{PARA 6 "" 1 "" 
{TEXT -1 6 "3! = 6" }}{PARA 6 "" 1 "" {TEXT -1 7 "4! = 24" }}{PARA 6 "
" 1 "" {TEXT -1 8 "5! = 120" }}{PARA 6 "" 1 "" {TEXT -1 8 "6! = 720" }
}{PARA 6 "" 1 "" {TEXT -1 9 "7! = 5040" }}{PARA 6 "" 1 "" {TEXT -1 10 
"8! = 40320" }}{PARA 6 "" 1 "" {TEXT -1 11 "9! = 362880" }}{PARA 6 "" 
1 "" {TEXT -1 13 "10! = 3628800" }}{PARA 6 "" 1 "" {TEXT -1 14 "11! = \+
39916800" }}{PARA 6 "" 1 "" {TEXT -1 15 "12! = 479001600" }}{PARA 6 "
" 1 "" {TEXT -1 16 "13! = 6227020800" }}{PARA 6 "" 1 "" {TEXT -1 17 "1
4! = 87178291200" }}{PARA 6 "" 1 "" {TEXT -1 19 "15! = 1307674368000" 
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 8 "Questi
on" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 73 "In how many ways c
an the 26 letters of the English alphabet be arranged? " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 8 "Solution" }{TEXT -1 
2 ": " }}{PARA 0 "" 0 "" {TEXT -1 26 "Sample arrangements are:  " }}
{PARA 256 "" 0 "" {TEXT -1 53 " A B C D E F G H IJ K L M N O P Q R S T
 U V W X Y Z, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" 
{TEXT -1 54 " Z Y X W V U T S R Q P O N M L K J I H G F E D C B A, " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 54 " Q U I
 C K B W N F O X J M P S V R T H E L A Z Y D G, " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 54 " N E W J O B F I X M R \+
G L U C K S H A Z Y T V P D Q. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 36 "The total number of arrangements is " }
{XPPEDIT 18 0 "26! = 26*`.`*25*`.`*24*` . . . `*3*`.`*2*`.`*1" "6#/-%*
factorialG6#\"#E*8F'\"\"\"%\".GF)\"#DF)F*F)\"#CF)%(~.~.~.~GF)\"\"$F)F*
F)\"\"#F)F*F)F)F)" }{XPPEDIT 18 0 "``=403291461126605635584000000" "6#
/%!G\"<+++%eNcgE6Y\"H.%" }{TEXT -1 1 " " }{TEXT 277 1 "~" }{TEXT -1 1 
" " }{XPPEDIT 18 0 "4*`.`*10^26" "6#*(\"\"%\"\"\"%\".GF%\"#5\"#E" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 32 "This is a pretty large n
umber!  " }}{PARA 0 "" 0 "" {TEXT -1 14 "A trillion is " }{XPPEDIT 18 
0 "10^12" "6#*$\"#5\"#7" }{TEXT -1 92 " (in the American system for la
rge numbers), so this last number is 400 trillion trillion.  " }}
{PARA 0 "" 0 "" {TEXT -1 64 "For comparison, the total number of atoms
 in your body is about " }{XPPEDIT 18 0 "10^28" "6#*$\"#5\"#G" }{TEXT 
-1 37 ", and the mass of the earth is about " }{XPPEDIT 18 0 "3.6*`.`*
10^27" "6#*(-%&FloatG6$\"#O!\"\"\"\"\"%\".GF)\"#5\"#F" }{TEXT -1 8 " g
rams. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 
"Note" }{TEXT -1 31 ": Arrangements are also called " }{TEXT 259 12 "p
ermutations" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "26!;\nevalf(%);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#\"<+++%eNcgE6Y\"H.%" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+6Y\"H.%\"#<" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 61 "A procedure for constructing arrangements of distinct i
tems: " }{TEXT 0 7 "arrange" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "The following procedure " }
{TEXT 0 7 "arrange" }{TEXT -1 87 " can be used to enumerate all the po
ssible arrangements of the objects in a Maple list." }}{PARA 15 "" 0 "
" {TEXT -1 46 "The objects in the list must all be different." }}
{PARA 15 "" 0 "" {TEXT -1 48 "There can be no more than 7 objects in t
he list." }}{PARA 15 "" 0 "" {TEXT -1 27 "The list must not be empty.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "The p
rocedure is defined in a " }{TEXT 259 17 "recursive fashion" }{TEXT 
-1 67 ", that is, the procedure makes use of itself within the definit
ion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 213 "
Once one knows how to arrange a given number of objects, it is straigh
tforward to explain how to get all the arrangements of a list containi
ng one more object. Remove one item and then arrange the remaining ite
ms." }}{PARA 0 "" 0 "" {TEXT -1 141 "Insert the chosen object before t
he objects in each of these arrangements. Repeat this for all possible
 choices of objects in the given list." }}{PARA 0 "" 0 "" {TEXT -1 
177 "To get things started it is necessary to show how to get all arra
ngements of a list containing a single object, which is easy to do sin
ce there is only one possible arrangement." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 798 "arrange := proc(i
tems::list)\n   local n,perms,i,j,item,partperm;\n   n := nops(items);
\n   if n=1 then return [items];\n   elif n>7 then\n      error \"the \+
list contains too many objects\"\n   elif n=0 then\n      error \"the \+
list must contain at least one object\"\n   end if;\n   if nops(\{op(i
tems)\})<>n then\n      error \"the list must contain distinct objects
\"\n   end if;\n   perms := NULL;\n   for i to n do\n      #Pick one i
tem.\n      item := items[i];\n  \n      # Permute the other items (Th
is is where the recursion occurs).\n      partperm := arrange([op(1..i
-1,items),op(i+1..n,items)]);\n      \n      # Form all arrangements i
n which the chosen item appears first.\n      for j from 1 to nops(par
tperm) do\n         perms := perms,[item,op(partperm[j])];\n      end \+
do;\n   end do;\n   return[perms];\nend proc:" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "arrange([A,B
]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$%\"AG%\"BG7$F&F%" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "ar
range([A,B,C]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%\"AG%\"BG%\"CG
7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "combinat[permute]([A,
A,B]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%7%%\"AGF%%\"BG7%F%F&F%7%F&
F%F%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "
We can obtain the number of members of a list by using the procedure \+
" }{TEXT 0 4 "nops" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "arrange([A,B,C,D]);\nnops(%)
;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"AG%\"BG%\"CG%\"DG7&F%F&F(F
'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F
(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F
%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F
&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#C" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "arrange([A
,B,C,D,E]);\nnops(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7dr7'%\"AG%\"
BG%\"CG%\"DG%\"EG7'F%F&F'F)F(7'F%F&F(F'F)7'F%F&F(F)F'7'F%F&F)F'F(7'F%F
&F)F(F'7'F%F'F&F(F)7'F%F'F&F)F(7'F%F'F(F&F)7'F%F'F(F)F&7'F%F'F)F&F(7'F
%F'F)F(F&7'F%F(F&F'F)7'F%F(F&F)F'7'F%F(F'F&F)7'F%F(F'F)F&7'F%F(F)F&F'7
'F%F(F)F'F&7'F%F)F&F'F(7'F%F)F&F(F'7'F%F)F'F&F(7'F%F)F'F(F&7'F%F)F(F&F
'7'F%F)F(F'F&7'F&F%F'F(F)7'F&F%F'F)F(7'F&F%F(F'F)7'F&F%F(F)F'7'F&F%F)F
'F(7'F&F%F)F(F'7'F&F'F%F(F)7'F&F'F%F)F(7'F&F'F(F%F)7'F&F'F(F)F%7'F&F'F
)F%F(7'F&F'F)F(F%7'F&F(F%F'F)7'F&F(F%F)F'7'F&F(F'F%F)7'F&F(F'F)F%7'F&F
(F)F%F'7'F&F(F)F'F%7'F&F)F%F'F(7'F&F)F%F(F'7'F&F)F'F%F(7'F&F)F'F(F%7'F
&F)F(F%F'7'F&F)F(F'F%7'F'F%F&F(F)7'F'F%F&F)F(7'F'F%F(F&F)7'F'F%F(F)F&7
'F'F%F)F&F(7'F'F%F)F(F&7'F'F&F%F(F)7'F'F&F%F)F(7'F'F&F(F%F)7'F'F&F(F)F
%7'F'F&F)F%F(7'F'F&F)F(F%7'F'F(F%F&F)7'F'F(F%F)F&7'F'F(F&F%F)7'F'F(F&F
)F%7'F'F(F)F%F&7'F'F(F)F&F%7'F'F)F%F&F(7'F'F)F%F(F&7'F'F)F&F%F(7'F'F)F
&F(F%7'F'F)F(F%F&7'F'F)F(F&F%7'F(F%F&F'F)7'F(F%F&F)F'7'F(F%F'F&F)7'F(F
%F'F)F&7'F(F%F)F&F'7'F(F%F)F'F&7'F(F&F%F'F)7'F(F&F%F)F'7'F(F&F'F%F)7'F
(F&F'F)F%7'F(F&F)F%F'7'F(F&F)F'F%7'F(F'F%F&F)7'F(F'F%F)F&7'F(F'F&F%F)7
'F(F'F&F)F%7'F(F'F)F%F&7'F(F'F)F&F%7'F(F)F%F&F'7'F(F)F%F'F&7'F(F)F&F%F
'7'F(F)F&F'F%7'F(F)F'F%F&7'F(F)F'F&F%7'F)F%F&F'F(7'F)F%F&F(F'7'F)F%F'F
&F(7'F)F%F'F(F&7'F)F%F(F&F'7'F)F%F(F'F&7'F)F&F%F'F(7'F)F&F%F(F'7'F)F&F
'F%F(7'F)F&F'F(F%7'F)F&F(F%F'7'F)F&F(F'F%7'F)F'F%F&F(7'F)F'F%F(F&7'F)F
'F&F%F(7'F)F'F&F(F%7'F)F'F(F%F&7'F)F'F(F&F%7'F)F(F%F&F'7'F)F(F%F'F&7'F
)F(F&F%F'7'F)F(F&F'F%7'F)F(F'F%F&7'F)F(F'F&F%" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"$?\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "arrange([A,B,C,D,E,F]):\nnops(%);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$?(" }}}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 "Note" }{TEXT -1 29 ": There is a M
aple procedure " }{TEXT 0 7 "permute" }{TEXT -1 30 " in the combinator
ics package " }{TEXT 0 8 "combinat" }{TEXT -1 41 " which can be used t
o list permutations. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 27 "combinat[permute]([A,B,C]);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#7(7%%\"AG%\"BG%\"CG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F
%F&7%F'F&F%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 42 "combinat[permute]([A,B,C,D,E,F]):\nnops(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$?(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "
" 0 "" {TEXT -1 24 "General permutations of " }{TEXT 278 1 "r" }{TEXT 
-1 20 " objects taken from " }{TEXT 279 2 "n " }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 72 "In some situat
ions we may not want to arrange all the objects available." }}{PARA 0 
"" 0 "" {TEXT -1 106 "For example, we may want to find how many ways t
here are of arranging 3 letters from the English alphabet." }}{PARA 0 
"" 0 "" {TEXT -1 147 "Since there are 26 letters altogether, there are
 26 choices for the first letter, 25 for the second letter and 24 for \+
the third, giving a total of " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "26*`.`*25*`.`*24 = 15600;" "6#/*,\"#E\"\"\"%\".GF&\"#DF
&F'F&\"#CF&\"&+c\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 147 "If we al
lowed the same letter to appear more than once, so that such possibili
ties as SOS or ADD are included, the total number of possibilities is
" }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "26*`.`*26*`.`*26
" "6#*,\"#E\"\"\"%\".GF%F$F%F&F%F$F%" }{TEXT -1 3 " = " }{XPPEDIT 18 
0 "26^3 = 17576;" "6#/*$\"#E\"\"$\"&wv\"" }{TEXT -1 2 ". " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "However, these pos
sibilities with repetitions are " }{TEXT 259 16 "not arrangements" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 53 "An arrangement of 3 lett
ers chosen from 26 is called " }{TEXT 259 45 "a permutation of 26 obje
cts taken 3 at a time" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 316 "Suppose that a courier has 7 package
s to deliver to 7 different addresses. The courier has 7 choices for t
he first package (and address). Once the first package has been delive
red there are 6 choices for the second package (and address), and so o
n. The total number of ways in which the packages can be delivered is \+
" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "7! = 7*`.`*6*`.`*
5*`.`*4*`.`*3*`.`*2*`.`*1;" "6#/-%*factorialG6#\"\"(*<F'\"\"\"%\".GF)
\"\"'F)F*F)\"\"&F)F*F)\"\"%F)F*F)\"\"$F)F*F)\"\"#F)F*F)F)F)" }
{XPPEDIT 18 0 "`` = 5040;" "6#/%!G\"%S]" }{TEXT -1 2 ". " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 150 "In practice the c
ourier would be likely choose a route which would attempt to minimise \+
the distance travelled, but this is not to be considered here. " }}
{PARA 0 "" 0 "" {TEXT -1 198 "Now suppose that 4 of the packages must \+
be delivered before lunch, but it does not matter which 4 from the 7 a
re delivered. The number of ways of delivering 4 packages taken from t
he 7 packages is  " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 
"7*`.`*6*`.`*5*`.`*4 = 840;" "6#/*0\"\"(\"\"\"%\".GF&\"\"'F&F'F&\"\"&F
&F'F&\"\"%F&\"$S)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 68 "Thi
s is the number of permutations of 7 packages taken 4 at a time. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "In genera
l, the " }{TEXT 259 26 "number of permutations of " }{TEXT 283 1 "r" }
{TEXT 259 20 " objects taken from " }{TEXT 284 1 "n" }{TEXT -1 4 " is \+
" }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "n*`.`*(n-1)*`.`*
(n-2)*` . . . `*(n-r+1)" "6#*0%\"nG\"\"\"%\".GF%,&F$F%F%!\"\"F%F&F%,&F
$F%\"\"#F(F%%(~.~.~.~GF%,(F$F%%\"rGF(F%F%F%" }{TEXT -1 2 ". " }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "A convenient ex
pression for this involving factorials is " }{XPPEDIT 18 0 "n!/(n-r)!;
" "6#*&-%*factorialG6#%\"nG\"\"\"-F%6#,&F'F(%\"rG!\"\"F-" }{TEXT -1 
48 ",  and this expression is denoted by the symbol " }{XPPEDIT 18 0 "
P(n,r)" "6#-%\"PG6$%\"nG%\"rG" }{TEXT -1 3 ",  " }{XPPEDIT 18 0 "P[n,r
]" "6#&%\"PG6$%\"nG%\"rG" }{TEXT -1 5 "  or " }{XPPEDIT 18 0 "``[n];" 
"6#&%!G6#%\"nG" }{XPPEDIT 18 0 "P[r];" "6#&%\"PG6#%\"rG" }{TEXT -1 1 "
." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "P[n,r] = n!/(n-r
)!;" "6#/&%\"PG6$%\"nG%\"rG*&-%*factorialG6#F'\"\"\"-F+6#,&F'F-F(!\"\"
F1" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 267 10 
"__________" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "For example, th
e number of permutations of 7 objects taken 4 at a time is " }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "P[7,4] = 7*`.`*6*`.`*5*`.`*
4*`.`*3*`.`*2*`.`*1/(3*`.`*2*`.`*1);" "6#/&%\"PG6$\"\"(\"\"%*>F'\"\"\"
%\".GF*\"\"'F*F+F*\"\"&F*F+F*F(F*F+F*\"\"$F*F+F*\"\"#F*F+F*F*F**,F.F*F
+F*F/F*F+F*F*F*!\"\"" }{XPPEDIT 18 0 "`` = 7*`.`*6*`.`*5*`.`*4;" "6#/%
!G*0\"\"(\"\"\"%\".GF'\"\"'F'F(F'\"\"&F'F(F'\"\"%F'" }{XPPEDIT 18 0 "`
` = 840;" "6#/%!G\"$S)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 83 "For another example, the number of p
ermutations of 26 letters taken 3 at a time is " }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "P[26,3] = 26!/23!;" "6#/&%\"PG6$\"#E\"
\"$*&-%*factorialG6#F'\"\"\"-F+6#\"#B!\"\"" }{XPPEDIT 18 0 "``=26*`.`*
25*`.`*24" "6#/%!G*,\"#E\"\"\"%\".GF'\"#DF'F(F'\"#CF'" }{XPPEDIT 18 0 
"``=15600" "6#/%!G\"&+c\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 "Note" }{TEXT -1 75 ": In order \+
that this expression gives the correct result in the case where " }
{XPPEDIT 18 0 "r=n" "6#/%\"rG%\"nG" }{TEXT -1 21 ", it is necessary to
 " }{TEXT 259 26 "define 0! to be equal to 1" }{TEXT -1 44 ", so that \+
the number of permutations of all " }{TEXT 315 1 "n" }{TEXT -1 14 " it
ems (taken " }{TEXT 316 1 "n" }{TEXT -1 16 " at a time) is: " }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "P[n,n]=n!/(n-n)!" "6#/&%\"P
G6$%\"nGF'*&-%*factorialG6#F'\"\"\"-F*6#,&F'F,F'!\"\"F0" }{XPPEDIT 18 
0 "``=n!/0!" "6#/%!G*&-%*factorialG6#%\"nG\"\"\"-F'6#\"\"!!\"\"" }
{XPPEDIT 18 0 "``=n!" "6#/%!G-%*factorialG6#%\"nG" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 20 "The Maple procedure " }{TEXT 0 8 "numbperm" }
{TEXT -1 16 " in the package " }{TEXT 0 8 "combinat" }{TEXT -1 29 " pe
rforms these calculations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 62 "The number of permutations of 7 packages taken 4 \+
at a time is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 24 "combinat[numbperm](7,4);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"$S)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 62 "The number of permutations of 26 letters taken 3 at a \+
time is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "combinat[numbperm](26,3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"&+c\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 34 "An example involving permutations " }}{PARA 0 "" 0 "" 
{TEXT 317 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 
171 "Find the number of different flags which can be made which consis
t of three different vertical coloured bands, where the possible colou
rs are red, green, blue, and yellow." }}{PARA 0 "" 0 "" {TEXT -1 67 "M
ake a list of the permutations which describe the possible flags. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 318 8 "Solution
" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 136 "There are 4 differe
nt colours which can be used for the first band leaving 3 possibilitie
s for the second band and 2 for the third band." }}{PARA 0 "" 0 "" 
{TEXT -1 29 "The total number of flags is " }}{PARA 256 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "4*`.`*3*`.`*2=24" "6#/*,\"\"%\"\"\"%\".GF&\"
\"$F&F'F&\"\"#F&\"#C" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 112 
"Note that this is the number of permutations of 4 items taken 3 at a \+
time. Hence the answer can be obtained from" }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "P[4,3]=4!/1!" "6#/&%\"PG6$\"\"%\"\"$*&-
%*factorialG6#F'\"\"\"-F+6#F-!\"\"" }{XPPEDIT 18 0 "``=24" "6#/%!G\"#C
" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 87 "We can get Maple to make a list of the possible colours u
sed in the bands of the flag. " }}{PARA 0 "" 0 "" {TEXT -1 137 "In the
 following list of permutations the colours red, green, blue and yello
w are represented by the letters R, G, B and Y respectively. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
58 "unassign('R','B','G','Y'):\ncombinat[permute]([R,G,B,Y],3);" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7%%\"RG%\"GG%\"BG7%F%F&%\"YG7%F%F'F&
7%F%F'F)7%F%F)F&7%F%F)F'7%F&F%F'7%F&F%F)7%F&F'F%7%F&F'F)7%F&F)F%7%F&F)
F'7%F'F%F&7%F'F%F)7%F'F&F%7%F'F&F)7%F'F)F%7%F'F)F&7%F)F%F&7%F)F%F'7%F)
F&F%7%F)F&F'7%F)F'F%7%F)F'F&" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 141 "The permutations of the colours used for
 the flags can be given without the square brackets and commas involve
d in the Maple list structure. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "flags := combinat[permute](
[R,G,B,Y],3):\nnops(flags);\nop(map(_V->convert(StringTools[Join](map(
_U->convert(_U,string),_V)),'name'),flags));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"#C" }}{PARA 12 "" 1 "" {XPPMATH 20 "6:%&R~G~BG%&R~G~Y
G%&R~B~GG%&R~B~YG%&R~Y~GG%&R~Y~BG%&G~R~BG%&G~R~YG%&G~B~RG%&G~B~YG%&G~Y
~RG%&G~Y~BG%&B~R~GG%&B~R~YG%&B~G~RG%&B~G~YG%&B~Y~RG%&B~Y~GG%&Y~R~GG%&Y
~R~BG%&Y~G~RG%&Y~G~BG%&Y~B~RG%&Y~B~GG" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 83 "The following code constructs an anim
ation which can be used to display the flags. " }}{PARA 0 "" 0 "" 
{TEXT -1 142 "It is probably best to display the flags one at a time u
sing the ->| control in the context bar (which appears when the graphi
c is selected). " }}{PARA 0 "" 0 "" {TEXT -1 112 "Alternatively, the a
nimation speed can be reduced to display the flags at a slow rate such
 as 2 frames per sec. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 572 "for i to nops(flags) do\n   for j \+
to 3 do\n      if op(j,flags[i])=R then clr[i,j] := COLOR(RGB,1,0,0)\n
      elif op(j,flags[i])=G then clr[i,j] := COLOR(RGB,0,1,0)\n      e
lif op(j,flags[i])=B then clr[i,j] := COLOR(RGB,0,0,1)\n      elif op(
j,flags[i])=Y then clr[i,j] := COLOR(RGB,1,1,0)\n      else error \"ca
nnot assign colour\" end if;\n   end do;\nend do:\nfrms := NULL:\nfor \+
i to nops(flags) do\n   pp := seq(plots[polygonplot]([[j,0],[j,1],[j+1
,1],[j+1,0]],color=clr[i,j]),j=1..3):\n   frms := frms,plots[display](
pp,axes=none);\nend do:\nplots[display]([frms],insequence=true);" }}
{PARA 13 "" 1 "" {GLPLOT2D 264 165 165 {PLOTDATA 2 "6#-%(ANIMATEG6:7&-
%)POLYGONSG6$7&7$$\"\"\"\"\"!$F.F.7$F,F,7$$\"\"#F.F,7$F2F/-%&COLORG6&%
$RGBGF,F/F/-F(6$7&F4F17$$\"\"$F.F,7$F=F/-F66&F8F/F,F/-F(6$7&F?F<7$$\"
\"%F.F,7$FFF/-F66&F8F/F/F,-%*AXESSTYLEG6#%%NONEG7&F'F9-F(6$FD-F66&F8F,
F,F/FK7&F'-F(6$F;FI-F(6$FDF@FK7&F'FUFPFK7&F'-F(6$F;FRFWFK7&F'FenFBFK7&
-F(6$F*F@-F(6$F;F5FBFK7&FinF[oFPFK7&FinFU-F(6$FDF5FK7&FinFUFPFK7&FinFe
nF_oFK7&FinFenFBFK7&-F(6$F*FIF[oFWFK7&FeoF[oFPFK7&FeoF9F_oFK7&FeoF9FPF
K7&FeoFenF_oFK7&FeoFenFWFK7&-F(6$F*FRF[oFWFK7&F]pF[oFBFK7&F]pF9F_oFK7&
F]pF9FBFK7&F]pFUF_oFK7&F]pFUFWFK" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
47.000000 38.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}
{SECT 1 {PARA 4 "" 0 "" {TEXT -1 26 "Using the Maple procedure " }
{TEXT 0 17 "combinat[permute]" }{TEXT -1 22 " to list permutations " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "The Mapl
e combinatorics package " }{TEXT 0 8 "combinat" }{TEXT -1 17 " has a p
rocedure " }{TEXT 0 7 "permute" }{TEXT -1 71 " for obtaining permutati
ons of the members of a list or set containing " }{TEXT 280 1 "n" }
{TEXT -1 23 " objects where we take " }{TEXT 281 1 "r" }{TEXT -1 43 " \+
objects at a time. Of course we must have " }{XPPEDIT 18 0 "r <= n;" "
6#1%\"rG%\"nG" }{TEXT -1 13 ". The number " }{TEXT 282 1 "r" }{TEXT 
-1 39 " is given as a second input parameter. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "The number of permutation
s of the four letters A, B, C and D, taken 2 at a time is " }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "P(4,2)=4!/2!" "6#/-%\"PG6$
\"\"%\"\"#*&-%*factorialG6#F'\"\"\"-F+6#F(!\"\"" }{XPPEDIT 18 0 "``=4*
`.`*3" "6#/%!G*(\"\"%\"\"\"%\".GF'\"\"$F'" }{XPPEDIT 18 0 "``=12" "6#/
%!G\"#7" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "combinat[permute]([A,B,C,D],2);\nno
ps(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7.7$%\"AG%\"BG7$F%%\"CG7$F%%
\"DG7$F&F%7$F&F(7$F&F*7$F(F%7$F(F&7$F(F*7$F*F%7$F*F&7$F*F(" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"#7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
107 "In this example there are 4 choices for the 1st object and 3 choi
ces for the 2nd object, making a total of " }{XPPEDIT 18 0 "4*`.`*3 = \+
12" "6#/*(\"\"%\"\"\"%\".GF&\"\"$F&\"#7" }{TEXT -1 30 " arrangements o
r permutations." }}{PARA 0 "" 0 "" {TEXT -1 101 "As another example, t
he number of permutations of the 5 letters A, B, C, D and E taken 3 at
 a time is" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "P(5,3)=
5!/2!" "6#/-%\"PG6$\"\"&\"\"$*&-%*factorialG6#F'\"\"\"-F+6#\"\"#!\"\"
" }{XPPEDIT 18 0 "`` = 5*`.`*4*`.`*3;" "6#/%!G*,\"\"&\"\"\"%\".GF'\"\"
%F'F(F'\"\"$F'" }{XPPEDIT 18 0 "``=60" "6#/%!G\"#g" }{TEXT -1 2 ". " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 42 "combinat[permute]([A,B,C,D,E],3);\nnops(%);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#7hn7%%\"AG%\"BG%\"CG7%F%F&%\"DG7%F%F&%\"EG7%F%F'F&7%F%F
'F)7%F%F'F+7%F%F)F&7%F%F)F'7%F%F)F+7%F%F+F&7%F%F+F'7%F%F+F)7%F&F%F'7%F
&F%F)7%F&F%F+7%F&F'F%7%F&F'F)7%F&F'F+7%F&F)F%7%F&F)F'7%F&F)F+7%F&F+F%7
%F&F+F'7%F&F+F)7%F'F%F&7%F'F%F)7%F'F%F+7%F'F&F%7%F'F&F)7%F'F&F+7%F'F)F
%7%F'F)F&7%F'F)F+7%F'F+F%7%F'F+F&7%F'F+F)7%F)F%F&7%F)F%F'7%F)F%F+7%F)F
&F%7%F)F&F'7%F)F&F+7%F)F'F%7%F)F'F&7%F)F'F+7%F)F+F%7%F)F+F&7%F)F+F'7%F
+F%F&7%F+F%F'7%F+F%F)7%F+F&F%7%F+F&F'7%F+F&F)7%F+F'F%7%F+F'F&7%F+F'F)7
%F+F)F%7%F+F)F&7%F+F)F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#g" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 26 "Selections or combinati
ons" }}{PARA 0 "" 0 "" {TEXT 285 8 "Question" }{TEXT -1 2 ": " }}
{PARA 0 "" 0 "" {TEXT -1 71 "In how many ways can we choose 3 people f
rom 5 to serve on a committee?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 286 8 "Solution" }{TEXT -1 4 ":   " }}{PARA 0 "
" 0 "" {TEXT -1 133 "We start by obtaining all possible arrangements o
f 3 people taken from 5. We designate the 5 people by the letters A, B
, C, D and E. " }}{PARA 0 "" 0 "" {TEXT -1 10 "There are " }{XPPEDIT 
18 0 "P(5,3)=5!/2!" "6#/-%\"PG6$\"\"&\"\"$*&-%*factorialG6#F'\"\"\"-F+
6#\"\"#!\"\"" }{XPPEDIT 18 0 "``=5*`.`*4*`.`*3" "6#/%!G*,\"\"&\"\"\"%
\".GF'\"\"%F'F(F'\"\"$F'" }{XPPEDIT 18 0 "``=60" "6#/%!G\"#g" }{TEXT 
-1 20 " such arrangements. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 85 " A B C, A B D, A B E, A C B, A C D, A C E, A D \+
B, A D C, A D E, A E B, A E C, A E D, " }}{PARA 0 "" 0 "" {TEXT -1 85 
" B A C, B A D, B A E, B C A, B C D, B C E, B D A, B D C, B D E, B E A
, B E C, B E D, " }}{PARA 0 "" 0 "" {TEXT -1 85 " C A B, C A D, C A E,
 C B A, C B D, C B E, C D A, C D B, C D E, C E A, C E B, C E D, " }}
{PARA 0 "" 0 "" {TEXT -1 85 " D A B, D A C, D A E, D B A, D B C, D B E
, D C A, D C B, D C E, D E A, D E B, D E C, " }}{PARA 0 "" 0 "" {TEXT 
-1 86 " E A B, E A C, E A D, E B A, E B C, E B D, E C A, E C B, E C D,
 E D A, E D B, E D C.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 42 "The first of these arrangements is A B C. " }}{PARA 
0 "" 0 "" {TEXT 259 89 "All 3! = 6 arrangements of these 3 objects amo
ng themselves are present in the above list" }{TEXT -1 1 "." }}{PARA 
0 "" 0 "" {TEXT -1 116 "These are A B C and A C B in the first line, B
 A C and B C A in the second line, C A B and C B A in the third line. \+
" }}{PARA 0 "" 0 "" {TEXT -1 102 "Since the order is not important for
 a selection, all these 6 arrangements really constitute just the " }
{TEXT 259 13 "one selection" }{TEXT -1 31 " of 3 people namely A, B an
d C." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "
Similarly, the selection of A, B and D is represented in the previous \+
list of 60 arrangements by the " }{XPPEDIT 18 0 "3!=6" "6#/-%*factoria
lG6#\"\"$\"\"'" }{TEXT -1 51 " arrangements of the three people A, B a
nd D among " }}{PARA 0 "" 0 "" {TEXT -1 12 "themselves. " }}{PARA 0 "
" 0 "" {TEXT -1 117 "These are A B D and A D B in the first line, B A \+
D and B D A in the second line, D A B and D B A in the fourth line. " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 388 "The 60
 arrangements of A, B, C, D and E can be grouped together in 10 groups
 of 6 such that each of the 10 groups determines just one selection in
 which the order is disregarded. These 10 groups and the corresponding
 selections are as follows. Each selection is given as a set designate
d by the curly brackets \{..\} in order to emphasise that the order wi
thin the selection is immaterial. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 59 " A B C, A C B, B A C, B C A, C A B, C B A
 ----- \{A, B, C\}  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 59 " A B D, A D B, B A D, B D A, D A B, D B A ----- \{A, B,
 D\}  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 
" A B E, A E B, B A E, B E A, E A B, E B A ----- \{A, B, E\}   " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 " A C D, A
 D C, C A D, C D A, D A C, D C A ----- \{A, C, D\}  " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 " A C E, A E C, C A E, C
 E A, E A C, E C A ----- \{A, C, E\}  " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 59 " A D E, A E D, D A E, D E A, E A D, E
 D A ----- \{A, D, E\}  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 59 " B C D, B D C, C B D, C D B, D B C, D C B ----- \{
B, C, D\}  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 59 " B C E, B E C, C B E, C E B, E B C, E C B ----- \{B, C, E\}  " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 " B D E,
 B E D, D B E, D E B, E B D, E D B ----- \{B, D, E\}  " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 " C D E, C E D, D C E
, D E C, E C D, E D C ----- \{C, D, E\} " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "The conclusion is that it is possi
ble to form " }{TEXT 261 22 "10 possible committees" }{TEXT -1 39 " co
nsisting of 3 people chosen from 5. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 225 "The number of arrangements of 3 people
 taken from 5 (of 5 people taken 3 at a time) can be recovered by mult
iplying the number of selections (namely 10) by the number of arrangem
ents of the 3 people in each selection (namely " }{XPPEDIT 18 0 "3!=6
" "6#/-%*factorialG6#\"\"$\"\"'" }{TEXT -1 41 "). Thus if the number o
f selections, say " }{TEXT 299 1 "N" }{TEXT -1 22 ", is unknown, we ha
ve " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "N*`.`*3! = P[5
,3];" "6#/*(%\"NG\"\"\"%\".GF&-%*factorialG6#\"\"$F&&%\"PG6$\"\"&F+" }
{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "
" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "N = P[5,3]/3!;" "6#/%\"NG*&&%\"P
G6$\"\"&\"\"$\"\"\"-%*factorialG6#F*!\"\"" }{XPPEDIT 18 0 "``=5!/(2!*`
.`*3!)" "6#/%!G*&-%*factorialG6#\"\"&\"\"\"*(-F'6#\"\"#F*%\".GF*-F'6#
\"\"$F*!\"\"" }{XPPEDIT 18 0 "``=5*`.`*4/(2*`.`*1)" "6#/%!G**\"\"&\"\"
\"%\".GF'\"\"%F'*(\"\"#F'F(F'F'F'!\"\"" }{XPPEDIT 18 0 "``=10" "6#/%!G
\"#5" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 
0 "" {TEXT -1 82 "The previous argument can be generalised. Suppose we
 want to determine the number " }{TEXT 300 1 "N" }{TEXT -1 27 " of pos
sible selections of " }{TEXT 288 1 "r" }{TEXT -1 19 " items chosen fro
m " }{TEXT 289 1 "n" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 30 "T
he number of arrangements of " }{TEXT 297 1 "r" }{TEXT -1 18 " items t
aken from " }{TEXT 298 1 "n" }{TEXT -1 4 " is " }{XPPEDIT 18 0 "P[n,r]
" "6#&%\"PG6$%\"nG%\"rG" }{TEXT -1 6 ". The " }{TEXT 293 1 "r" }{TEXT 
-1 71 " items in any particular selection can be arranged among themse
lves in " }{XPPEDIT 18 0 "r!" "6#-%*factorialG6#%\"rG" }{TEXT -1 7 " w
ays. " }}{PARA 0 "" 0 "" {TEXT -1 13 "If there are " }{TEXT 296 1 "N" 
}{TEXT -1 17 " selections, the " }{XPPEDIT 18 0 "P[n,r]" "6#&%\"PG6$%
\"nG%\"rG" }{TEXT -1 43 " arrangements can be grouped together into " 
}{TEXT 295 1 "N" }{TEXT -1 25 " groups, each containing " }{XPPEDIT 
18 0 "r!" "6#-%*factorialG6#%\"rG" }{TEXT -1 15 " arrangements. " }}
{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+
" }{XPPEDIT 18 0 "N*`.`*r!=P[n,r]" "6#/*(%\"NG\"\"\"%\".GF&-%*factoria
lG6#%\"rGF&&%\"PG6$%\"nGF+" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT 
-1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "N=P
[n,r]/r!" "6#/%\"NG*&&%\"PG6$%\"nG%\"rG\"\"\"-%*factorialG6#F*!\"\"" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 58 "A selection of objects in no particular order is called a
 " }{TEXT 259 11 "combination" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 12 "In general, " }{TEXT 259 31 "the number of ways of choosi
ng " }{TEXT 290 1 "r" }{TEXT 259 14 " objects from " }{TEXT 291 1 "n" 
}{TEXT 259 8 " objects" }{TEXT -1 41 " (without regard to order) is de
noted by " }{XPPEDIT 18 0 "C(n,r)" "6#-%\"CG6$%\"nG%\"rG" }{TEXT -1 3 
",  " }{XPPEDIT 18 0 "C[n,r]" "6#&%\"CG6$%\"nG%\"rG" }{TEXT -1 5 "  or
 " }{XPPEDIT 18 0 "``[n];" "6#&%!G6#%\"nG" }{XPPEDIT 18 0 "C[r];" "6#&
%\"CG6#%\"rG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 20 "This is \+
also called " }{TEXT 259 30 "the number of combinations of " }{TEXT 
294 1 "n" }{TEXT 259 13 " items taken " }{TEXT 292 1 "r" }{TEXT 259 
10 " at a time" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "C[n,r]=P[n,r]/r!" "6#/&%\"CG6$%\"nG%\"rG*&&%\"PG6$F'F(
\"\"\"-%*factorialG6#F(!\"\"" }{XPPEDIT 18 0 "`` = n!/(r!*(n-r)!);" "6
#/%!G*&-%*factorialG6#%\"nG\"\"\"*&-F'6#%\"rGF*-F'6#,&F)F*F.!\"\"F*F2
" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 287 15 "_
______________" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 90 "For example, the number of ways of select
ing 5 cards from a standard deck of 52 cards is: " }}{PARA 256 "" 0 "
" {TEXT -1 1 " " }{XPPEDIT 18 0 "C[52,5] = 52!/(5!*`.`*47!);" "6#/&%\"
CG6$\"#_\"\"&*&-%*factorialG6#F'\"\"\"*(-F+6#F(F-%\".GF--F+6#\"#ZF-!\"
\"" }{XPPEDIT 18 0 "``=52*`.`*51*`.`*50*`.`*49*`.`*48/(5*`.`*4*`.`*3*`
.`*2*`.`*1)" "6#/%!G*6\"#_\"\"\"%\".GF'\"#^F'F(F'\"#]F'F(F'\"#\\F'F(F'
\"#[F'*4\"\"&F'F(F'\"\"%F'F(F'\"\"$F'F(F'\"\"#F'F(F'F'F'!\"\"" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "``=2598960" "6#/%!G\"(g*)f#" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "For anot
her example, the number of ways that an artist can chose 6 paintings f
rom a possible 9 to place in an exhibition is:  " }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "C[9, 6] = 9!/(6!*`.`*3!);" "6#/&%\"CG6$
\"\"*\"\"'*&-%*factorialG6#F'\"\"\"*(-F+6#F(F-%\".GF--F+6#\"\"$F-!\"\"
" }{XPPEDIT 18 0 "`` = 9*`.`*8*`.`*7/(3*`.`*2*`.`*1);" "6#/%!G*.\"\"*
\"\"\"%\".GF'\"\")F'F(F'\"\"(F'*,\"\"$F'F(F'\"\"#F'F(F'F'F'!\"\"" }
{XPPEDIT 18 0 "`` = 84;" "6#/%!G\"#%)" }{TEXT -1 2 ". " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Whenever, " }{TEXT 
309 1 "r" }{TEXT -1 25 " items are selected from " }{TEXT 310 1 "n" }
{TEXT -1 18 " items, there are " }{XPPEDIT 18 0 "n-r" "6#,&%\"nG\"\"\"
%\"rG!\"\"" }{TEXT -1 92 " items left behind. This observation leads t
o the fact that the number of ways of selecting " }{TEXT 311 1 "r" }
{TEXT -1 12 " items from " }{TEXT 312 1 "n" }{TEXT -1 65 " is the same
 as the number of ways of selecting (leaving behind) " }{XPPEDIT 18 0 
"n-r" "6#,&%\"nG\"\"\"%\"rG!\"\"" }{TEXT -1 12 " items from " }{TEXT 
313 1 "n" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}
{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "C[n,r]=C[n,n-r]" "6#/
&%\"CG6$%\"nG%\"rG&F%6$F',&F'\"\"\"F(!\"\"" }{TEXT -1 2 ". " }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{TEXT 314 9 "_________" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "Although \+
this follows logically by the previous explanation, it also follows al
gebraically. " }}{PARA 0 "" 0 "" {TEXT -1 16 "On the one hand " }}
{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "C[n,r]=n!/(r!*(n-r)!)
" "6#/&%\"CG6$%\"nG%\"rG*&-%*factorialG6#F'\"\"\"*&-F+6#F(F--F+6#,&F'F
-F(!\"\"F-F4" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 26 "while, o
n the other hand, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 
"C[n,n-r]=n!/((n-r)!*`.`*((n-(n-r))!)" "6#/&%\"CG6$%\"nG,&F'\"\"\"%\"r
G!\"\"*&-%*factorialG6#F'F)*(-F.6#,&F'F)F*F+F)%\".GF)-F.6#,&F'F),&F'F)
F*F+F+F)F+" }{XPPEDIT 18 0 "``=n!/((n-r)!*r!)" "6#/%!G*&-%*factorialG6
#%\"nG\"\"\"*&-F'6#,&F)F*%\"rG!\"\"F*-F'6#F/F*F0" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "For examp
le, the number of ways of selecting 6 paintings from 9 is " }{XPPEDIT 
18 0 "C[9,6]=84" "6#/&%\"CG6$\"\"*\"\"'\"#%)" }{TEXT -1 58 ", and the \+
number of ways of selecting 3 pictures from 9 is" }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "C[9,3]=9!/(3!*`.`*6!)" "6#/&%\"CG6$\"\"
*\"\"$*&-%*factorialG6#F'\"\"\"*(-F+6#F(F-%\".GF--F+6#\"\"'F-!\"\"" }
{XPPEDIT 18 0 "``=84" "6#/%!G\"#%)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 6 "also. " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT -1 20 "The Maple procedure " }{TEXT 0 8 "numbcomb" }{TEXT -1 
16 " in the package " }{TEXT 0 8 "combinat" }{TEXT -1 29 " performs th
ese calculations." }}{PARA 0 "" 0 "" {TEXT -1 51 "The number of ways o
f selecting 5 cards from 52 is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "combinat[numbcomb](52,5);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"(g*)f#" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "The number of ways of selecting
 47 cards from 52 is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 26 "combinat[numbcomb](52,47);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#\"(g*)f#" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 64 "The number of ways of chosing 6 paintings
 from a possible 9 is: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "combinat[numbcomb](9,6);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"#%)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 53 "The number of ways of chosing 3 paintings
 from 9 is: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 24 "combinat[numbcomb](9,3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"#%)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 
46 "Maple procedures connected with combinations: " }{TEXT 0 12 "combi
neperms" }{TEXT -1 2 ", " }{TEXT 0 12 "permstocombs" }{TEXT -1 2 ", " 
}{TEXT 0 16 "combinat[choose]" }{TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 16 "The pr
ocedures: " }{TEXT 0 12 "combineperms" }{TEXT -1 2 ", " }{TEXT 0 12 "p
ermstocombs" }{TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 54 "Given a list of arrangements, th
e following procedure " }{TEXT 0 12 "combineperms" }{TEXT -1 144 " wil
l group together all those arrangements which involve the same objects
. In this way it provides a reduction from arrangements to selections.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 616 "combineperms := proc(permutations::listlist)\n   local j,n,gr
oups,p,perms,q,subgroup;\n   perms := permutations;\n   groups := NULL
;\n   n := nops(perms);\n   while n>1 do\n      p := perms[1];\n      \+
perms := [op(2..nops(perms),perms)];\n      n := n-1;\n      j := 1;\n
      subgroup := p;\n      while j<=n do\n         q := perms[j];\n  \+
       if \{op(q)\}=\{op(p)\} then\n            perms := [op(1..j-1,pe
rms),op(j+1..n,perms)];\n            subgroup := subgroup,q;\n        \+
    n := n-1;\n         else\n            j := j+1;\n         end if;
\n      end do;\n      groups := groups,[subgroup];\n   end do;\n   re
turn[groups];\nend proc:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 19 "The next procedure " }{TEXT 0 12 "permstocombs" }
{TEXT -1 163 " is similar to the previous procedure except that each o
f the groups is replaced by a single set to represent the associated c
ombination. As its name suggests, it " }{TEXT 259 37 "converts permuta
tions to combinations" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 523 "permstocombs := proc(pe
rmutations::listlist)\n   local j,n,combs,p,perms;\n   perms := permut
ations;\n   combs := NULL:\n   n := nops(perms);\n   while n>1 do\n   \+
   p := perms[1];\n      combs := combs,\{op(p)\};\n      perms := [op
(2..nops(perms),perms)];\n      n := n-1;\n      j := 1;\n      while \+
j<=n do\n         if \{op(perms[j])\}=\{op(p)\} then\n            perm
s := [op(1..j-1,perms),op(j+1..n,perms)];\n            n := n-1;\n    \+
     else\n            j := j+1;\n         end if;\n      end do;\n   \+
end do;\n   return[combs];\nend proc:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 114 "Consider the problem of determining the number of
 ways in which 3 people be chosen from 5 to serve on a committee?" }}
{PARA 0 "" 0 "" {TEXT -1 84 "As before, we start by obtaining all poss
ible arrangements of 3 people taken from 5." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "perms := combinat[
permute]([A,B,C,D,E],3);\nnops(perms);" }}{PARA 12 "" 1 "" {XPPMATH 
20 "6#>%&permsG7hn7%%\"AG%\"BG%\"CG7%F'F(%\"DG7%F'F(%\"EG7%F'F)F(7%F'F
)F+7%F'F)F-7%F'F+F(7%F'F+F)7%F'F+F-7%F'F-F(7%F'F-F)7%F'F-F+7%F(F'F)7%F
(F'F+7%F(F'F-7%F(F)F'7%F(F)F+7%F(F)F-7%F(F+F'7%F(F+F)7%F(F+F-7%F(F-F'7
%F(F-F)7%F(F-F+7%F)F'F(7%F)F'F+7%F)F'F-7%F)F(F'7%F)F(F+7%F)F(F-7%F)F+F
'7%F)F+F(7%F)F+F-7%F)F-F'7%F)F-F(7%F)F-F+7%F+F'F(7%F+F'F)7%F+F'F-7%F+F
(F'7%F+F(F)7%F+F(F-7%F+F)F'7%F+F)F(7%F+F)F-7%F+F-F'7%F+F-F(7%F+F-F)7%F
-F'F(7%F-F'F)7%F-F'F+7%F-F(F'7%F-F(F)7%F-F(F+7%F-F)F'7%F-F)F(7%F-F)F+7
%F-F+F'7%F-F+F(7%F-F+F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#g" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "The first arrangement in this l
ist is " }{XPPEDIT 18 0 "[A,B,C]" "6#7%%\"AG%\"BG%\"CG" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT 259 82 "All arrangements of these 3 objects \+
among themselves are present in the above list" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 61 "We can check this as follows. First find \+
all arrangements of " }{TEXT 301 1 "A" }{TEXT -1 2 ", " }{TEXT 302 1 "
B" }{TEXT -1 5 " and " }{TEXT 303 1 "C" }{TEXT -1 2 ". " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "sub := \+
combinat[permute]([A,B,C]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$subG
7(7%%\"AG%\"BG%\"CG7%F'F)F(7%F(F'F)7%F(F)F'7%F)F'F(7%F)F(F'" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "Now check using
 intersection of sets that all these arrangements are present." }}
{PARA 0 "" 0 "" {TEXT -1 67 "Note that the lists must be converted to \+
sets before the operation " }{TEXT 0 9 "intersect" }{TEXT -1 13 " can \+
be used." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "\{op(sub)\} intersect \{op(perm)\};" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#<(7%%\"AG%\"BG%\"CG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7
%F'F&F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 "Since the order is \+
not important for a selection, all these 6 arrangements really constit
ute just the one selection of 3 people namely " }{TEXT 304 1 "A" }
{TEXT -1 2 ", " }{TEXT 305 1 "B" }{TEXT -1 5 " and " }{TEXT 306 1 "C" 
}{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 83 "The 2nd arrangement among the 60 arrangements of the 5 pe
ople taken 3 at a time is " }{XPPEDIT 18 0 "[A,B,D]" "6#7%%\"AG%\"BG%
\"DG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 111 "Again we find th
at all the arrangements of these 3 people among themselves are include
d in the 60 arrangements." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "combinat[permute]([A,B,D]);\n\{op(%
)\} intersect \{op(perm)\};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%\"
AG%\"BG%\"DG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#<(7%%\"AG%\"BG%\"DG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7
%F'F&F%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
14 "The procedure " }{TEXT 0 12 "combineperms" }{TEXT -1 92 " can be u
sed to group together all the arrangements associated with a particula
r selection. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 74 "perms := combinat[permute]([A,B,C,D,E],3):\ncomb
ineperms(perms);\ngrps := %:" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7,7(7%
%\"AG%\"BG%\"CG7%F&F(F'7%F'F&F(7%F'F(F&7%F(F&F'7%F(F'F&7(7%F&F'%\"DG7%
F&F0F'7%F'F&F07%F'F0F&7%F0F&F'7%F0F'F&7(7%F&F'%\"EG7%F&F8F'7%F'F&F87%F
'F8F&7%F8F&F'7%F8F'F&7(7%F&F(F07%F&F0F(7%F(F&F07%F(F0F&7%F0F&F(7%F0F(F
&7(7%F&F(F87%F&F8F(7%F(F&F87%F(F8F&7%F8F&F(7%F8F(F&7(7%F&F0F87%F&F8F07
%F0F&F87%F0F8F&7%F8F&F07%F8F0F&7(7%F'F(F07%F'F0F(7%F(F'F07%F(F0F'7%F0F
'F(7%F0F(F'7(7%F'F(F87%F'F8F(7%F(F'F87%F(F8F'7%F8F'F(7%F8F(F'7(7%F'F0F
87%F'F8F07%F0F'F87%F0F8F'7%F8F'F07%F8F0F'7(7%F(F0F87%F(F8F07%F0F(F87%F
0F8F(7%F8F(F07%F8F0F(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 78 "The individual groups can be printed on separate lin
es by the following loop. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "for i to nops(grps) do print(grps[i
]) end do;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%\"AG%\"BG%\"CG7%F%F
'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#7(7%%\"AG%\"BG%\"DG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#7(7%%\"AG%\"BG%\"EG7%F%F'F&7%F&F%F'7%F&F'F%7
%F'F%F&7%F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%\"AG%\"CG%\"DG
7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#7(7%%\"AG%\"CG%\"EG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%\"AG%\"DG%\"EG7%F%F'F&7%F&F%F'7%F
&F'F%7%F'F%F&7%F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%\"BG%\"C
G%\"DG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7(7%%\"BG%\"CG%\"EG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'
F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%\"BG%\"DG%\"EG7%F%F'F&7%F
&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(7%%
\"CG%\"DG%\"EG7%F%F'F&7%F&F%F'7%F&F'F%7%F'F%F&7%F'F&F%" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 157 "Alternatively, the \+
following code displays the groups as individual rows of a matrix, and
 also removes all the commas involved in the Maple list structures. " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 111 "map(_W->map(_V->convert(StringTools[Join](map(_U->convert(_U,st
ring),_V)),'name'),_W),grps):\nconvert(%,matrix);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#K%'matrixG6#7,7(%&A~B~CG%&A~C~BG%&B~A~CG%&B~C~AG%&C~A~B
G%&C~B~AG7(%&A~B~DG%&A~D~BG%&B~A~DG%&B~D~AG%&D~A~BG%&D~B~AG7(%&A~B~EG%
&A~E~BG%&B~A~EG%&B~E~AG%&E~A~BG%&E~B~AG7(%&A~C~DG%&A~D~CG%&C~A~DG%&C~D
~AG%&D~A~CG%&D~C~AG7(%&A~C~EG%&A~E~CG%&C~A~EG%&C~E~AG%&E~A~CG%&E~C~AG7
(%&A~D~EG%&A~E~DG%&D~A~EG%&D~E~AG%&E~A~DG%&E~D~AG7(%&B~C~DG%&B~D~CG%&C
~B~DG%&C~D~BG%&D~B~CG%&D~C~BG7(%&B~C~EG%&B~E~CG%&C~B~EG%&C~E~BG%&E~B~C
G%&E~C~BG7(%&B~D~EG%&B~E~DG%&D~B~EG%&D~E~BG%&E~B~DG%&E~D~BG7(%&C~D~EG%
&C~E~DG%&D~C~EG%&D~E~CG%&E~C~DG%&E~D~CGQ(pprint36\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 198 "The 60 arrangements of 5 people taken 3 \+
at a time can be grouped together in 10 groups of 6 where each group o
f 6 arrangements just involves arranging a given selection of 3 people
 amomg themselves." }}{PARA 0 "" 0 "" {TEXT -1 116 "Each of the 10 gro
ups of arrangements gives rise to just one selection in which we ignor
e the order of the 3 people." }}{PARA 0 "" 0 "" {TEXT -1 65 "The numbe
r of ways of choosing  3 people from 5 is therefore 10. " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "Since the order o
f the objects in a selection or combination is not important, we can r
epresent a selection by means of a " }{TEXT 259 3 "set" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 73 "Thus when using Maple we can represen
t permutations (or arrangements) by " }{TEXT 259 5 "lists" }{TEXT -1 
37 " and combinations (or selections) by " }{TEXT 259 4 "sets" }{TEXT 
-1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
14 "The procedure " }{TEXT 0 12 "permstocombs" }{TEXT -1 38 " is simil
ar to the previous procedure " }{TEXT 0 12 "combineperms" }{TEXT -1 
126 " except that each of the groups is replaced by a single set to re
present the associated combination. As its name suggests, it " }{TEXT 
259 37 "converts permutations to combinations" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
72 "perms := combinat[permute]([A,B,C,D,E],3):\npermstocombs(perms);\n
nops(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7,<%%\"AG%\"CG%\"BG<%%\"DG
F%F'<%F%%\"EGF'<%F)F%F&<%F%F&F+<%F)F%F+<%F)F&F'<%F&F+F'<%F)F+F'<%F)F&F
+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#5" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 46 "There are 10 ways of choosing 3 people from 5." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "The Maple
 procedure " }{TEXT 0 8 "numbcomb" }{TEXT -1 16 " in the package " }
{TEXT 0 8 "combinat" }{TEXT -1 53 " can be used to obtain the number o
f ways of chosing " }{TEXT 307 1 "r" }{TEXT -1 12 " items from " }
{TEXT 308 1 "n" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 51 "The nu
mber of ways of selecting 3 people from 7 is:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "combinat[num
bcomb](5,3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#5" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 88 "For another example, consider selecting 4
 people from the 6 people A, B, C, D, E, and F." }}{PARA 0 "" 0 "" 
{TEXT -1 38 "The number of possible selections is: " }}{PARA 256 "" 0 
"" {TEXT -1 1 " " }{XPPEDIT 18 0 "C[6,4] = 6!/(4!*`.`*2!);" "6#/&%\"CG
6$\"\"'\"\"%*&-%*factorialG6#F'\"\"\"*(-F+6#F(F-%\".GF--F+6#\"\"#F-!\"
\"" }{XPPEDIT 18 0 "``=(6*`.`*5)/(2*`.`*1)" "6#/%!G**\"\"'\"\"\"%\".GF
'\"\"&F'*(\"\"#F'F(F'F'F'!\"\"" }{XPPEDIT 18 0 "``=15" "6#/%!G\"#:" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 86 "Consider the arrangements of 4 people from 6, that is, of
 6 people taken 4 at a time. " }}{PARA 0 "" 0 "" {TEXT -1 127 "These a
rrangements can be grouped together so that each group is associated w
ith just one selection, and involves the possible " }{XPPEDIT 18 0 "4!
=24" "6#/-%*factorialG6#\"\"%\"#C" }{TEXT -1 49 " arrangements of the \+
4 people in each selection. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 15 "The procedures " }{TEXT 0 12 "combineperms" }
{TEXT -1 5 " and " }{TEXT 0 12 "permstocombs" }{TEXT -1 60 " are used \+
in the following code segment to illustrate this. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "perms := co
mbinat[permute]([A,B,C,D,E,F],4):\ngrps := combineperms(perms):\nselec
tions := permstocombs(perms):\nfor i to nops(grps) do print(selections
[i]);print(grps[i]);print(``); end do;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#<&%\"DG%\"AG%\"CG%\"BG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%
\"AG%\"BG%\"CG%\"DG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7
&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7
&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7
&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"AG%\"CG%\"EG%\"BG" }
}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"AG%\"BG%\"CG%\"EG7&F%F&F(F'7&F
%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F
&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F
'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F
(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#<&%\"FG%\"AG%\"CG%\"BG" }}{PARA 12 "" 1 "" {XPPMATH 20 
"6#7:7&%\"AG%\"BG%\"CG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F
%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F
'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F
(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"DG%\"AG%\"E
G%\"BG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"AG%\"BG%\"DG%\"EG7&F%
F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&
F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'
F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(
F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#<&%\"DG%\"FG%\"AG%\"BG" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#7:7&%\"AG%\"BG%\"DG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7
&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7
&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7
&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"
FG%\"AG%\"EG%\"BG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"AG%\"BG%\"
EG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F
&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F
'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F
(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"DG%\"AG%\"CG%\"EG" }}{PARA 12 "" 
1 "" {XPPMATH 20 "6#7:7&%\"AG%\"CG%\"DG%\"EG7&F%F&F(F'7&F%F'F&F(7&F%F'
F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(
F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(
F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#<&%\"DG%\"FG%\"AG%\"CG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"AG%
\"CG%\"DG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%
F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%
F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&
F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"FG%\"AG%\"CG%\"EG" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"AG%\"CG%\"EG%\"FG7&F%F&F(F'7&F%
F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&
F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'
F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(
F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#<&%\"DG%\"FG%\"AG%\"EG" }}{PARA 12 "" 1 "" {XPPMATH 20 
"6#7:7&%\"AG%\"DG%\"EG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F
%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F
'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F
(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"DG%\"CG%\"E
G%\"BG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"BG%\"CG%\"DG%\"EG7&F%
F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&
F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'
F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(
F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#<&%\"DG%\"FG%\"CG%\"BG" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#7:7&%\"BG%\"CG%\"DG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7
&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7
&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7
&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"
FG%\"CG%\"EG%\"BG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"BG%\"CG%\"
EG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F
&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F
'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F
(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#<&%\"DG%\"FG%\"EG%\"BG" }}{PARA 12 "" 
1 "" {XPPMATH 20 "6#7:7&%\"BG%\"DG%\"EG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'
F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(
F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(
F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#<&%\"DG%\"FG%\"CG%\"EG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7:7&%\"CG%
\"DG%\"EG%\"FG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F%F(F'F&7&F&F%
F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F'F%F&F(7&F'F%
F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F(F%F'F&7&F(F&
F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
%!G" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "combinat[numbcomb](6,4);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"#:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 20 "The Maple procedure " }{TEXT 0 6 "choose" }{TEXT -1 16 
" in the package " }{TEXT 0 8 "combinat" }{TEXT -1 37 " can be used to
 enumerate selections." }}{PARA 0 "" 0 "" {TEXT -1 85 "The original ob
jects from which the selection is to be made should be given as a set.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 43 "combinat[choose](\{A,B,C,D,E,F\},4);\nnops(%);" }}{PARA 12 "" 
1 "" {XPPMATH 20 "6#<1<&%\"DG%\"CG%\"EG%\"BG<&%\"FG%\"AGF&F'<&F%F*F+F&
<&F%F*F+F'<&F*F&F'F(<&F%F*F&F(<&F%F*F'F(<&F%F*F&F'<&F%F+F&F'<&F%F*F+F(
<&F+F&F'F(<&F%F+F&F(<&F*F+F'F(<&F*F+F&F(<&F%F+F'F(" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#\"#:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" 
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 213 "The n
umber of ways of selecting 2 people from 6 is also 15. With each selec
tion of 2 people, 4 are left behind. In both the cases of selecting ei
ther of 2 or 4 people, we are really looking at the various ways of " 
}{TEXT 259 12 "partitioning" }{TEXT -1 120 " the original set into two
 mutually disjoint subsets whose union is the original set.  These par
titions are as follows. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 136 "S := \{A,B,C,D,E,F\}:\nselections \+
:= combinat[choose](S,2):\nfor i to nops(selections) do print(selectio
ns[i],S minus selections[i]) end do;" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"DG%\"FG<&%\"EG%\"AG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"DG%\"EG<&%\"FG%\"AG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"DG%\"AG<&%\"FG%\"EG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"DG%\"BG<&%\"FG%\"EG%\"AG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"DG%\"CG<&%\"FG%\"EG%\"AG%\"BG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"FG%\"EG<&%\"DG%\"AG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"FG%\"AG<&%\"DG%\"EG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"FG%\"BG<&%\"DG%\"EG%\"AG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"FG%\"CG<&%\"DG%\"EG%\"AG%\"BG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"EG%\"AG<&%\"DG%\"FG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"EG%\"BG<&%\"DG%\"FG%\"AG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"EG%\"CG<&%\"DG%\"FG%\"AG%\"BG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"AG%\"BG<&%\"DG%\"FG%\"EG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"AG%\"CG<&%\"DG%\"FG%\"EG%\"BG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<$%\"BG%\"CG<&%\"DG%\"FG%\"EG%\"AG" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Tasks" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT 
-1 3 "Q1 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
287 "Jack's cafe provides a special lunch deal with a choice of sandwi
ch, desert and a drink. There are three types of sandwich offered: ham
, salami and tuna, two deserts: brownie or Nanaimo bar, and four choic
es of drink: tea, coffee, coke or pepsi. How many different lunches ar
e possible? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "
" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "3
*`.`*2*`.`*4=24" "6#/*,\"\"$\"\"\"%\".GF&\"\"#F&F'F&\"\"%F&\"#C" }
{TEXT -1 1 " " }}}{PARA 0 "" 0 "" {TEXT -1 35 "_______________________
____________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 35 "___________________________________" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q2
 " }}{PARA 0 "" 0 "" {TEXT -1 145 "In how many ways can a multiple cho
ice test be answered if it consists of 10 questions each of which has \+
5 choices a, b, c, d, e for the answer? " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" 
{TEXT -1 10 "There are " }{XPPEDIT 18 0 "5^10=9765625" "6#/*$\"\"&\"#5
\"(Dcw*" }{TEXT -1 32 " ways to answer the whole test. " }}}{PARA 0 "
" 0 "" {TEXT -1 35 "___________________________________" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 35 "____________
_______________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";
" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q3 " }}{PARA 0 "" 0 "" 
{TEXT -1 218 "How many different license plates are possible if each p
late has three digits followed by three letters? Assume that repetitio
n of numbers or letters is allowed and that any of the 10 digits or 26
 letters may be used. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 5 "" 0 "" {TEXT -1 3 "Ans" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "10^3*`.`*26^3 = 17576000;" "6#/*(\"#5\"\"$%\".G\"\"\"\"
#EF&\")+gd<" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 
0 "" 0 "" {TEXT -1 35 "___________________________________" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 35 "____________
_______________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";
" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q4 " }}{PARA 0 "" 0 "" 
{TEXT -1 376 "Lex Luther fires a nuclear missile into the San Andreas \+
Fault in California, causing a huge earthquake. During the resulting c
haos Superman has to save Lois Lane from a rock slide, rescue Jimmy fr
om a burst dam, stop a train from derailing, and stop a school bus fro
m plummeting from the Golden Gate Bridge. In how many ways can the \"m
an of steel\" perform these four rescues? " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "4!=24" "6#/-%*factorialG6#\"\"%\"#C" }
{TEXT -1 2 ". " }}}{PARA 0 "" 0 "" {TEXT -1 35 "______________________
_____________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 35 "___________________________________" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q5
 " }}{PARA 0 "" 0 "" {TEXT -1 177 "Find the number of different flags \+
which can be made which consist of three different vertical coloured b
ands, where the possible colours are red, green, blue, yellow and blac
k." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT 
-1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "P[5,3]=5!
/2!" "6#/&%\"PG6$\"\"&\"\"$*&-%*factorialG6#F'\"\"\"-F+6#\"\"#!\"\"" }
{XPPEDIT 18 0 "``=5*`.`*4*`.`*3" "6#/%!G*,\"\"&\"\"\"%\".GF'\"\"%F'F(F
'\"\"$F'" }{XPPEDIT 18 0 "``=60" "6#/%!G\"#g" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "We can ge
t Maple to make a list of the possible colours used in the bands of th
e flag. " }}{PARA 0 "" 0 "" {TEXT -1 140 "In the following list of per
mutations the colours red, green, blue and yellow are represented by t
he letters R, G, B, Y and W respectively. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 172 "unassign('R','G',
'B','Y','W'):\nflags := combinat[permute]([R,G,B,Y,W],3):\nnops(flags)
;\nop(map(_V->convert(StringTools[Join](map(_U->convert(_U,string),_V)
),'name'),flags));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#g" }}{PARA 
12 "" 1 "" {XPPMATH 20 "6hn%&R~G~BG%&R~G~YG%&R~G~WG%&R~B~GG%&R~B~YG%&R
~B~WG%&R~Y~GG%&R~Y~BG%&R~Y~WG%&R~W~GG%&R~W~BG%&R~W~YG%&G~R~BG%&G~R~YG%
&G~R~WG%&G~B~RG%&G~B~YG%&G~B~WG%&G~Y~RG%&G~Y~BG%&G~Y~WG%&G~W~RG%&G~W~B
G%&G~W~YG%&B~R~GG%&B~R~YG%&B~R~WG%&B~G~RG%&B~G~YG%&B~G~WG%&B~Y~RG%&B~Y
~GG%&B~Y~WG%&B~W~RG%&B~W~GG%&B~W~YG%&Y~R~GG%&Y~R~BG%&Y~R~WG%&Y~G~RG%&Y
~G~BG%&Y~G~WG%&Y~B~RG%&Y~B~GG%&Y~B~WG%&Y~W~RG%&Y~W~GG%&Y~W~BG%&W~R~GG%
&W~R~BG%&W~R~YG%&W~G~RG%&W~G~BG%&W~G~YG%&W~B~RG%&W~B~GG%&W~B~YG%&W~Y~R
G%&W~Y~GG%&W~Y~BG" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 83 "The following code constructs an animation which can be u
sed to display the flags. " }}{PARA 0 "" 0 "" {TEXT -1 142 "It is prob
ably best to display the flags one at a time using the ->| control in \+
the context bar (which appears when the graphic is selected). " }}
{PARA 0 "" 0 "" {TEXT -1 112 "Alternatively, the animation speed can b
e reduced to display the flags at a slow rate such as 2 frames per sec
. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 634 "for i to nops(flags) do\n   for j to 3 do\n      if \+
op(j,flags[i])=R then clr[i,j] := COLOR(RGB,1,0,0)\n      elif op(j,fl
ags[i])=G then clr[i,j] := COLOR(RGB,0,1,0)\n      elif op(j,flags[i])
=B then clr[i,j] := COLOR(RGB,0,0,1)\n      elif op(j,flags[i])=Y then
 clr[i,j] := COLOR(RGB,1,1,0)\n      elif op(j,flags[i])=W then clr[i,
j] := COLOR(RGB,1,1,1)\n      else error \"cannot assign colour\" end \+
if;\n   end do;\nend do:\nfrms := NULL:\nfor i to nops(flags) do\n   p
p := seq(plots[polygonplot]([[j,0],[j,1],[j+1,1],[j+1,0]],color=clr[i,
j]),j=1..3):\n   frms := frms,plots[display](pp,axes=none);\nend do:\n
plots[display]([frms],insequence=true);" }}{PARA 13 "" 1 "" {GLPLOT2D 
264 184 184 {PLOTDATA 2 "6#-%(ANIMATEG6hn7&-%)POLYGONSG6$7&7$$\"\"\"\"
\"!$F.F.7$F,F,7$$\"\"#F.F,7$F2F/-%&COLORG6&%$RGBGF,F/F/-F(6$7&F4F17$$
\"\"$F.F,7$F=F/-F66&F8F/F,F/-F(6$7&F?F<7$$\"\"%F.F,7$FFF/-F66&F8F/F/F,
-%*AXESSTYLEG6#%%NONEG7&F'F9-F(6$FD-F66&F8F,F,F/FK7&F'F9-F(6$FD-F66&F8
F,F,F,FK7&F'-F(6$F;FI-F(6$FDF@FK7&F'FZFPFK7&F'FZFUFK7&F'-F(6$F;FRFfnFK
7&F'F[oFBFK7&F'F[oFUFK7&F'-F(6$F;FWFfnFK7&F'F`oFBFK7&F'F`oFPFK7&-F(6$F
*F@-F(6$F;F5FBFK7&FeoFgoFPFK7&FeoFgoFUFK7&FeoFZ-F(6$FDF5FK7&FeoFZFPFK7
&FeoFZFUFK7&FeoF[oF\\pFK7&FeoF[oFBFK7&FeoF[oFUFK7&FeoF`oF\\pFK7&FeoF`o
FBFK7&FeoF`oFPFK7&-F(6$F*FIFgoFfnFK7&FgpFgoFPFK7&FgpFgoFUFK7&FgpF9F\\p
FK7&FgpF9FPFK7&FgpF9FUFK7&FgpF[oF\\pFK7&FgpF[oFfnFK7&FgpF[oFUFK7&FgpF`
oF\\pFK7&FgpF`oFfnFK7&FgpF`oFPFK7&-F(6$F*FRFgoFfnFK7&FeqFgoFBFK7&FeqFg
oFUFK7&FeqF9F\\pFK7&FeqF9FBFK7&FeqF9FUFK7&FeqFZF\\pFK7&FeqFZFfnFK7&Feq
FZFUFK7&FeqF`oF\\pFK7&FeqF`oFfnFK7&FeqF`oFBFK7&-F(6$F*FWFgoFfnFK7&FcrF
goFBFK7&FcrFgoFPFK7&FcrF9F\\pFK7&FcrF9FBFK7&FcrF9FPFK7&FcrFZF\\pFK7&Fc
rFZFfnFK7&FcrFZFPFK7&FcrF[oF\\pFK7&FcrF[oFfnFK7&FcrF[oFBFK" 1 2 0 1 
10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "C
urve 3" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 35 "__________________
_________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 35 "___________________________________" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q6
 " }}{PARA 0 "" 0 "" {TEXT -1 199 "Find the number of different flags \+
which can be made which consist of three different vertical coloured b
ands where the possible colours are red, green, blue, cyan, yellow, ma
genta, black and white. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "P[8,3] = 8!/5!;" "6#/&%\"PG6$\"\")\"\"$*&-%*factorialG6
#F'\"\"\"-F+6#\"\"&!\"\"" }{XPPEDIT 18 0 "`` = 8*`.`*7*`.`*6;" "6#/%!G
*,\"\")\"\"\"%\".GF'\"\"(F'F(F'\"\"'F'" }{XPPEDIT 18 0 "`` = 336;" "6#
/%!G\"$O$" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "combinat[numbperm](8,3);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"$O$" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT 
-1 35 "___________________________________" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 35 "_______________________________
____" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 
4 "" 0 "" {TEXT -1 3 "Q7 " }}{PARA 0 "" 0 "" {TEXT -1 142 "A new Toyot
a, a new Hyundai and a used Mercedes are to be assigned to three of te
n salespersons. In how many ways can the assignment be made? " }}
{PARA 0 "" 0 "" {TEXT -1 86 "Hint: Think of assigning the salespersons
 to the cars instead of the other way round. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "
" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "P[10,3]=10!/7!" "6#/&%\"PG6$\"#5
\"\"$*&-%*factorialG6#F'\"\"\"-F+6#\"\"(!\"\"" }{XPPEDIT 18 0 "``=10*`
.`*9*`.`*8" "6#/%!G*,\"#5\"\"\"%\".GF'\"\"*F'F(F'\"\")F'" }{XPPEDIT 
18 0 "``=720" "6#/%!G\"$?(" }{TEXT -1 2 ". " }}}{PARA 0 "" 0 "" {TEXT 
-1 35 "___________________________________" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 35 "_______________________________
____" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 
4 "" 0 "" {TEXT -1 3 "Q8 " }}{PARA 0 "" 0 "" {TEXT -1 90 "(a) Find the
 number of ways of choosing 3 people from the 7 people A, B, C, D, E, \+
F and G." }}{PARA 0 "" 0 "" {TEXT -1 34 "     List the possible select
ions." }}{PARA 0 "" 0 "" {TEXT -1 90 "(b) Find the number of ways of c
hoosing 4 people from the 7 people A, B, C, D, E, F and G." }}{PARA 0 
"" 0 "" {TEXT -1 34 "     List the possible selections." }}{PARA 0 "" 
0 "" {TEXT -1 58 "(c) Why are the numerical answers to (a) and (b) the
 same?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" 
{TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 4 "(a) " }{XPPEDIT 18 0 "
C[7,3]=7!/(3!*4!)" "6#/&%\"CG6$\"\"(\"\"$*&-%*factorialG6#F'\"\"\"*&-F
+6#F(F--F+6#\"\"%F-!\"\"" }{XPPEDIT 18 0 "``=7*`.`*6*`.`*5/(3*`.`*2*`.
`*1)" "6#/%!G*.\"\"(\"\"\"%\".GF'\"\"'F'F(F'\"\"&F'*,\"\"$F'F(F'\"\"#F
'F(F'F'F'!\"\"" }{XPPEDIT 18 0 "``=35" "6#/%!G\"#N" }{TEXT -1 2 ". " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "(b) " }
{XPPEDIT 18 0 "C[7,4] = 7!/(4!*3!);" "6#/&%\"CG6$\"\"(\"\"%*&-%*factor
ialG6#F'\"\"\"*&-F+6#F(F--F+6#\"\"$F-!\"\"" }{XPPEDIT 18 0 "``=7*`.`*6
*`.`*5/(3*`.`*2*`.`*1)" "6#/%!G*.\"\"(\"\"\"%\".GF'\"\"'F'F(F'\"\"&F'*
,\"\"$F'F(F'\"\"#F'F(F'F'F'!\"\"" }{XPPEDIT 18 0 "``=35" "6#/%!G\"#N" 
}{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 137 "(c) In either case the set \{A, B, C, D, E, F, G\} is pa
rtitioned into two subsets: one containing 3 members and one containin
g 4 members. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 54 "Maple can list the various combinations as subsets of " }
{XPPEDIT 18 0 "\{A, B, C, D, E, F, G\};" "6#<)%\"AG%\"BG%\"CG%\"DG%\"E
G%\"FG%\"GG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "combinat[choose](\{A,B,C,D,E
,F,G\},3);\nnops(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#<E<%%\"DG%\"FG
%\"EG<%F%F&%\"AG<%F%F&%\"BG<%F%F&%\"CG<%F%F&%\"GG<%F%F'F)<%F%F'F+<%F%F
'F-<%F%F'F/<%F%F)F+<%F%F)F-<%F%F)F/<%F%F+F-<%F%F+F/<%F%F-F/<%F&F'F)<%F
&F'F+<%F&F'F-<%F&F'F/<%F&F)F+<%F&F)F-<%F&F)F/<%F&F+F-<%F&F+F/<%F&F-F/<
%F'F)F+<%F'F)F-<%F'F)F/<%F'F+F-<%F'F+F/<%F'F-F/<%F)F+F-<%F)F+F/<%F)F-F
/<%F+F-F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#N" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "combinat[cho
ose](\{A,B,C,D,E,F,G\},4);\nnops(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "
6#<E<&%\"DG%\"AG%\"BG%\"GG<&%\"EGF&F'F(<&F%%\"FGF*F&<&F%F,F*F'<&F%F,F*
%\"CG<&F%F,F*F(<&F%F,F&F'<&F%F,F&F/<&F%F,F&F(<&F%F,F'F/<&F%F,F'F(<&F%F
,F/F(<&F%F*F&F'<&F%F*F&F/<&F%F*F&F(<&F%F*F'F/<&F%F*F'F(<&F%F*F/F(<&F%F
&F'F/<&F%F&F/F(<&F%F'F/F(<&F,F*F&F'<&F,F*F&F/<&F,F*F&F(<&F,F*F'F/<&F,F
*F'F(<&F,F*F/F(<&F,F&F'F/<&F,F&F'F(<&F,F&F/F(<&F,F'F/F(<&F*F&F'F/<&F*F
&F/F(<&F*F'F/F(<&F&F'F/F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#N" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "The two m
ethods of selection amount to the following partitions. " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138 "S :=
 \{A,B,C,D,E,F,G\}:\nselections := combinat[choose](S,3):\nfor i to no
ps(selections) do print(selections[i],S minus selections[i]) end do;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"FG%\"EG<&%\"AG%\"BG%\"CG%
\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"FG%\"AG<&%\"EG%\"BG%
\"CG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"FG%\"BG<&%\"EG%
\"AG%\"CG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"FG%\"CG<&%
\"EG%\"AG%\"BG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"FG%\"
GG<&%\"EG%\"AG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"
EG%\"AG<&%\"FG%\"BG%\"CG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"
DG%\"EG%\"BG<&%\"FG%\"AG%\"CG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$
<%%\"DG%\"EG%\"CG<&%\"FG%\"AG%\"BG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6$<%%\"DG%\"EG%\"GG<&%\"FG%\"AG%\"BG%\"CG" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6$<%%\"DG%\"AG%\"BG<&%\"FG%\"EG%\"CG%\"GG" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6$<%%\"DG%\"AG%\"CG<&%\"FG%\"EG%\"BG%\"GG" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"AG%\"GG<&%\"FG%\"EG%\"BG%\"CG" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"BG%\"CG<&%\"FG%\"EG%\"AG%\"G
G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"BG%\"GG<&%\"FG%\"EG%\"A
G%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"DG%\"CG%\"GG<&%\"FG%\"E
G%\"AG%\"BG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"FG%\"EG%\"AG<&%\"D
G%\"BG%\"CG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"FG%\"EG%\"BG<
&%\"DG%\"AG%\"CG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"FG%\"EG%
\"CG<&%\"DG%\"AG%\"BG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"FG%
\"EG%\"GG<&%\"DG%\"AG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%
\"FG%\"AG%\"BG<&%\"DG%\"EG%\"CG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<%%\"FG%\"AG%\"CG<&%\"DG%\"EG%\"BG%\"GG" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6$<%%\"FG%\"AG%\"GG<&%\"DG%\"EG%\"BG%\"CG" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6$<%%\"FG%\"BG%\"CG<&%\"DG%\"EG%\"AG%\"GG" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6$<%%\"FG%\"BG%\"GG<&%\"DG%\"EG%\"AG%\"CG" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"FG%\"CG%\"GG<&%\"DG%\"EG%\"AG%\"B
G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"EG%\"AG%\"BG<&%\"DG%\"FG%\"C
G%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"EG%\"AG%\"CG<&%\"DG%\"F
G%\"BG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"EG%\"AG%\"GG<&%\"D
G%\"FG%\"BG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"EG%\"BG%\"CG<
&%\"DG%\"FG%\"AG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"EG%\"BG%
\"GG<&%\"DG%\"FG%\"AG%\"CG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%\"EG%
\"CG%\"GG<&%\"DG%\"FG%\"AG%\"BG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$<%%
\"AG%\"BG%\"CG<&%\"DG%\"FG%\"EG%\"GG" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6$<%%\"AG%\"BG%\"GG<&%\"DG%\"FG%\"EG%\"CG" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6$<%%\"AG%\"CG%\"GG<&%\"DG%\"FG%\"EG%\"BG" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6$<%%\"BG%\"CG%\"GG<&%\"DG%\"FG%\"EG%\"AG" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" 
}}}}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 37 "_____________________________________" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 
"Q9 " }}{PARA 0 "" 0 "" {TEXT -1 81 "(a) Find the number of ways of ch
oosing 2 cards from a standard deck of 52 cards." }}{PARA 0 "" 0 "" 
{TEXT -1 81 "(b) Find the number of ways of choosing 4 cards from a st
andard deck of 52 cards." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 4 "(a) " 
}{XPPEDIT 18 0 "C[52,2]=52!/(2!*`.`*50!)" "6#/&%\"CG6$\"#_\"\"#*&-%*fa
ctorialG6#F'\"\"\"*(-F+6#F(F-%\".GF--F+6#\"#]F-!\"\"" }{XPPEDIT 18 0 "
``=52*`.`*51/(2*`.`*1)" "6#/%!G**\"#_\"\"\"%\".GF'\"#^F'*(\"\"#F'F(F'F
'F'!\"\"" }{XPPEDIT 18 0 "``=1326" "6#/%!G\"%E8" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "(b) " }
{XPPEDIT 18 0 "C[52,4]=52!/(4!*`.`*48!)" "6#/&%\"CG6$\"#_\"\"%*&-%*fac
torialG6#F'\"\"\"*(-F+6#F(F-%\".GF--F+6#\"#[F-!\"\"" }{XPPEDIT 18 0 "`
`=52*`.`*51*`.`*50*`.`*49/(4*`.`*3*`.`*2*`.`*1)" "6#/%!G*2\"#_\"\"\"%
\".GF'\"#^F'F(F'\"#]F'F(F'\"#\\F'*0\"\"%F'F(F'\"\"$F'F(F'\"\"#F'F(F'F'
F'!\"\"" }{XPPEDIT 18 0 "``=270725" "6#/%!G\"'D2F" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "combinat[numbcomb](52,2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"%E8
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "combinat[numbcomb](52,4);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"'D2F" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 37 "
_____________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "____________
_________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "
;" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 6 "Q10   " }}{PARA 0 "" 0 "" 
{TEXT -1 160 "Theresa has the prerequisites for six different computer
 science courses. In how many ways can she select three of these cours
es for her schedule next semester." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "C[6, 3] = 6!/(3!*`.`*3!);" "6#/&%\"CG6$\"\"'\"\"
$*&-%*factorialG6#F'\"\"\"*(-F+6#F(F-%\".GF--F+6#F(F-!\"\"" }{XPPEDIT 
18 0 "`` = 6*`.`*5*`.`*4/(3*`.`*2*`.`*1);" "6#/%!G*.\"\"'\"\"\"%\".GF'
\"\"&F'F(F'\"\"%F'*,\"\"$F'F(F'\"\"#F'F(F'F'F'!\"\"" }{XPPEDIT 18 0 "`
` = 20;" "6#/%!G\"#?" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "combinat[numbcomb](6,3);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#?" }}}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "
" {TEXT -1 35 "___________________________________" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 35 "____________________
_______________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 
{PARA 0 "" 0 "" {TEXT -1 23 "Code for pictures etc. " }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 21 
"Code for tree diagram" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 779 "p1 := plot([[[0,0],[1,1]],[[0,0],[
1,3]],\n[[0,0],[1,-1]],[[0,0],[1,-3]],[[1,3],[2,3.5]],[[1,3],[2,2.5]],
\n[[1,1],[2,1.5]],[[1,1],[2,0.5]],[[1,-3],[2,-3.5]],[[1,-3],[2,-2.5]],
\n[[1,-1],[2,-1.5]],[[1,-1],[2,-0.5]]],color=COLOR(RGB,0,0,0)):\np2 :=
 plot([[[0,0],[1,3],[1,1],[1,-1],[1,-3],[2,3.5],[2,2.5],\n[2,1.5],[2,.
5],[2,-.5],[2,-1.5],[2,-2.5],[2,-3.5]]$4],style=point,\n symbol=[circl
e,diamond,cross,circle],symbolsize=[10$3,12],\n color=[red$3,black]):
\nt1 := plots[textplot]([[.8,1.9,`gold`],[.7,.5,`white`],[.7,-.3,`gree
n`],\n  [.9,-2,`maroon`],[1.5,3.6,`manual`],[1.5,2.5,`automatic`],\n[1
.5,1.6,`manual`],[1.5,.5,`automatic`],[1.5,-.4,`manual`],\n[1.5,-1.5,`
automatic`],[1.5,-2.4,`manual`],[1.5,-3.5,`automatic`]],\n  color=COLO
R(RGB,.01,.01,.01)):\nplots[display]([p1,p2,t1],axes=none);" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 61 "Code o
btaining permutations in a simpler form (not as lists) " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 153 "perm :
= combinat[permute]([A,B,C,D,E],3):\ncombineperms(perm):\nop(10,%):\no
p(map(_V->convert(StringTools[Join](map(_U->convert(_U,string),_V)),'n
ame'),%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(%&C~D~EG%&C~E~DG%&D~C~EG
%&D~E~CG%&E~C~DG%&E~D~CG" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "perm
s := combinat[permute]([A,B,C,D,E],3):\nnops(perms);\nop(map(_V->conve
rt(StringTools[Join](map(_U->convert(_U,string),_V)),'name'),perms));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#g" }}{PARA 12 "" 1 "" {XPPMATH 
20 "6hn%&A~B~CG%&A~B~DG%&A~B~EG%&A~C~BG%&A~C~DG%&A~C~EG%&A~D~BG%&A~D~C
G%&A~D~EG%&A~E~BG%&A~E~CG%&A~E~DG%&B~A~CG%&B~A~DG%&B~A~EG%&B~C~AG%&B~C
~DG%&B~C~EG%&B~D~AG%&B~D~CG%&B~D~EG%&B~E~AG%&B~E~CG%&B~E~DG%&C~A~BG%&C
~A~DG%&C~A~EG%&C~B~AG%&C~B~DG%&C~B~EG%&C~D~AG%&C~D~BG%&C~D~EG%&C~E~AG%
&C~E~BG%&C~E~DG%&D~A~BG%&D~A~CG%&D~A~EG%&D~B~AG%&D~B~CG%&D~B~EG%&D~C~A
G%&D~C~BG%&D~C~EG%&D~E~AG%&D~E~BG%&D~E~CG%&E~A~BG%&E~A~CG%&E~A~DG%&E~B
~AG%&E~B~CG%&E~B~DG%&E~C~AG%&E~C~BG%&E~C~DG%&E~D~AG%&E~D~BG%&E~D~CG" }
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}}{MARK "4 \+
0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }