{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "Blue Emphasis" -1 256 "Times" 0 0 0 0 255 1 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "Green Emphasis" -1 257 "Times" 1 12 0 128 0 1 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "Maroon Emphasis" -1 258 "Times" 1 12 128 0 128 1 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "Purple Emphasis" -1 259 "Times" 1 12 102 0 230 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "Red Emphasis" -1 260 "Times " 1 12 255 0 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "Dark Red Emphasis" -1 261 "Times" 1 12 128 0 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 258 262 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" 261 263 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" 260 264 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 258 265 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE " " 261 266 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" 258 267 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" 261 268 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" 260 269 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 260 270 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 260 271 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 258 272 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" 261 273 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times " 1 18 0 0 128 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 3 0 3 0 2 2 0 1 } {PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 128 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Text Output" -1 6 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 2 2 2 2 2 1 2 1 3 1 } 1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Warning" -1 7 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Map le Output" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Bullet Item" -1 15 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 3 3 1 0 1 0 2 2 15 2 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 35 "An introduction to counting metho ds" }}{PARA 0 "" 0 "" {TEXT -1 37 "by Peter Stone, Nanaimo, B.C., Cana da" }}{PARA 0 "" 0 "" {TEXT -1 18 "Version: 19.8.2007" }}{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 38 " \+ can be used to illustrate the answer." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{GLPLOT2D 264 260 260 {PLOTDATA 2 "61-%'CURVESG6/7$7$\"\"!F(7$\" \"\"F*7$F'7$F*\"\"$7$F'7$F*!\"\"7$F'7$F*!\"$7$F,7$\"\"#$\"#NF07$F,7$F6 $\"#DF07$F)7$F6$\"#:F07$F)7$F6$\"\"&F07$F27$F6$!#NF07$F27$F6$!#DF07$F/ 7$F6$!#:F07$F/7$F6$!\"&F0-%&COLORG6&%$RGBGF(F(F(-%'POINTSG60F'F,F)F/F2 F5F:F>FBFRFNFJFF-%'SYMBOLG6#%'CIRCLEG-%%TEXTG6$7$$\"\")F0$\"#>F0%%gold G-F[o6$7$$\"\"(F0FC%&whiteG-F[o6$7$Ffo$F3F0%&greenG-F[o6$7$$\"\"*F0$! \"#F(%'maroonG-F[o6$7$F?$\"#OF0%'manualG-F[o6$7$F?F;%*automaticG-F[o6$ 7$F?$\"#;F0F[q-F[o6$7$F?FCF_q-F[o6$7$F?$!\"%F0F[q-F[o6$7$F?FOF_q-F[o6$ 7$F?$!#CF0F[q-F[o6$7$F?FGF_q-%*AXESSTYLEG6#%%NONEG" 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" "Curve 12" "Curve 13" "Curve 14" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "The basic " }{TEXT 259 19 "mult iplication 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 260 6 "r ways" }{TEXT 258 51 ", and a second c hoice can be made independently in " }{TEXT 260 6 "s ways" }{TEXT 258 49 ", then the two choices can be made in a total of " }{TEXT 260 10 " r . s ways" }{TEXT -1 1 "." }}{PARA 256 "" 0 "" {TEXT 264 28 "________ ____________________" }}{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 89 "In how many ways can Albert, Betsy , Chris and \+ David line up to have their picture 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 60 "Let A, B, C and D represent Albert, Betsy , Chris and David." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 15 "" 0 "" {TEXT -1 8 "We have " }{TEXT 259 1 "4" }{TEXT -1 78 " choices for the person in the left-hand or 1st posi tion, namely A,B, C and D." }}{PARA 15 "" 0 "" {TEXT -1 59 "Once the1s t position on the left has been filled there are " }{TEXT 259 1 "3" } {TEXT -1 214 " ways of choosing someone for the 2nd position to the ri ght of the1st position. For example, if C occupies the 1st position on the left, then we could choose A, B or D for the second position. Alt ogether there are " }{XPPEDIT 18 0 "4*`.`*3 = 12" "6#/*(\"\"%\"\"\"%\" .GF&\"\"$F&\"#7" }{TEXT -1 41 " ways of filling the first two position s." }}{PARA 15 "" 0 "" {TEXT -1 92 " Once the first two positions on t he left are filled, we have two people left, so there are " }{TEXT 256 1 "2" }{TEXT -1 55 " ways of filling the 3rd position. There are t herefore " }{XPPEDIT 18 0 "12*`.`*2 = 24" "6#/*(\"#7\"\"\"%\".GF&\"\"# F&\"#C" }{TEXT -1 43 " ways of filling the first three positions." }} {PARA 15 "" 0 "" {TEXT -1 112 "The last position on the right must the n be occupied by the remaining person, that is, it can be filled in on ly " }{TEXT 256 1 "1" }{TEXT -1 5 " way." }}{PARA 15 "" 0 "" {TEXT -1 31 "The four people can line up in " }{XPPEDIT 18 0 "4*`.`*3*`.`*2*`.` *1 = 24" "6#/*0\"\"%\"\"\"%\".GF&\"\"$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 are:" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 35 "ABCD, ABDC, ACBD, ACDB, ADBC, ADCB," }}{PARA 0 "" 0 "" {TEXT -1 35 "BACD, BADC, BCAD, BCDA, BDAC, BDCA," }} {PARA 0 "" 0 "" {TEXT -1 35 "CABD, CADB, CBAD, CBDA, CDAB, CDBA," }} {PARA 0 "" 0 "" {TEXT -1 34 "DABC, DACB, DBAC, DBCA, DCAB, DCBA" }} {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 because the positions from left to right \+ can be filled in 5 ways, 4 ways, 3 ways, 2 ways and 1 way respectively ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "It i s convenient to use the " }{TEXT 259 9 "factorial" }{TEXT -1 59 " nota tion 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#/-%*factorialG 6#%\"nG*8F'\"\"\"%#.~GF),&F'F)F)!\"\"F)%\".GF),&F'F)\"\"#F,F)%(~.~.~.~ GF)\"\"$F)F-F)F/F)F-F)F)F)" }{TEXT -1 18 ". ( n! is read as " }{TEXT 259 11 "n factorial" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "1! = 1" "6#/-%*factorialG6#\"\"\"F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "2! = 2*`.`*1" "6#/-%* factorialG6#\"\"#*(F'\"\"\"%\".GF)F)F)" }{TEXT -1 5 " = 2 " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "3! = 3*`.`*2*`.`* 1" "6#/-%*factorialG6#\"\"$*, 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)\"\"$F)F*F)\"\"#F)F*F)F)F)" }{TEXT -1 6 " = 24 " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "5! = 5*`.`*4*`.`*3*`.`*2*`.`*1;" "6#/-%*facto rialG6#\"\"&*4F'\"\"\"%\".GF)\"\"%F)F*F)\"\"$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 "14! = 87178291 200" }}{PARA 6 "" 1 "" {TEXT -1 19 "15! = 1307674368000" }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 272 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 73 "In how many ways can the 26 l etters of the English alphabet be arranged? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 273 8 "Solution" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 14 "The answer is " }{XPPEDIT 18 0 "26! = 26* `.`*25*`.`*24*` . . . `*3*`.`*2*`.`*1" "6#/-%*factorialG6#\"#E*8F'\"\" \"%\".GF)\"#DF)F*F)\"#CF)%(~.~.~.~GF)\"\"$F)F*F)\"\"#F)F*F)F)F)" } {TEXT -1 31 ", which is a pretty big number." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "26!;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#\"<+++%eNcgE6Y\"H.%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalf(26!);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+6Y\"H.%\"#<" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "The following p rocedure " }{TEXT 0 7 "arrange" }{TEXT -1 87 " can be used to enumerat e all the possible arrangements of the objects in a Maple list." }} {PARA 15 "" 0 "" {TEXT -1 46 "The objects in the list must all be diff erent." }}{PARA 15 "" 0 "" {TEXT -1 48 "There can be no more than 7 ob jects in the 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 procedure is defined in a " }{TEXT 259 17 "recursive fashion" }{TEXT -1 67 ", that is, the procedure makes use of itself within the \+ definition." }}{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 straightforward to explain how to get all the arrangements of a list \+ containing one more object. Remove one item and then arrange the remai ning items." }}{PARA 0 "" 0 "" {TEXT -1 141 "Insert the chosen object \+ before the 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 arrangements of a list containing a single object, which is easy \+ to do since there is only one possible arrangement." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 813 "arrange : = proc(items::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(items)\}) <> n then\n error \"the list must contain \+ distinct objects\"\n end if;\n perms := NULL;\n for i from 1 to \+ n do\n #Pick one item.\n item := items[i];\n \n # Perm ute the other items (This is where the recursion occurs).\n partp erm := arrange([op(1..i-1,items),op(i+1..n,items)]);\n \n # \+ Form all arrangements in which the chosen item appears first.\n f or j from 1 to nops(partperm) do\n perms := perms,[item,op(par tperm[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 "arrange([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 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "We can ob tain 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#\"$?(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 24 "General permutations of " }{TEXT 275 1 "r" } {TEXT -1 20 " objects taken from " }{TEXT 276 1 "n" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 33 "A fancy na me for arrangements is " }{TEXT 259 12 "permutations" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 72 "In some situations we may not want to a rrange all the objects available." }}{PARA 0 "" 0 "" {TEXT -1 107 " Fo r example, we may want to find how many ways there are of arranging 3 \+ letters from the English alphabet." }}{PARA 0 "" 0 "" {TEXT -1 147 "Si nce there are 26 letters altogether, there are 26 choices for the firs t letter, 25 for the second letter and 24 for the third, giving a tota l of " }{XPPEDIT 18 0 "26*`.`*25*`.`*24" "6#*,\"#E\"\"\"%\".GF%\"#DF%F &F%\"#CF%" }{TEXT -1 9 " = 15600." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "26*25*24;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#\"&+c\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 147 "If we allowed the same letter to appear more than once, so that such possibilities as SOS or ADD are included, the tota l number of possibilities is" }}{PARA 256 "" 0 "" {TEXT -1 2 " " } {XPPEDIT 18 0 "26*`.`*26*`.`*26" "6#*,\"#E\"\"\"%\".GF%F$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 50 "However, these possi bilities with repetitions are " }{TEXT 259 16 "not arrangements" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "26^3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"&wv\" " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "An a rrangement of 3 letters chosen from 26 is called " }{TEXT 259 45 "a pe rmutation of 26 objects taken 3 at a time" }{TEXT -1 1 "." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "The Maple combinat orics package " }{TEXT 0 8 "combinat" }{TEXT -1 17 " has a procedure \+ " }{TEXT 0 7 "permute" }{TEXT -1 139 " for obtaining permutations of t he members of a list or set containing n objects where we take r objec ts at a time. Of course we must have " }{XPPEDIT 18 0 "r <= n;" "6#1% \"rG%\"nG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "with(combinat):\npermute([A,B]);" } }{PARA 7 "" 1 "" {TEXT -1 67 "Warning, the protected name Chi has been redefined and unprotected\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$% \"AG%\"BG7$F&F%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "If you give the procedure " }{TEXT 0 7 "permute" }{TEXT -1 139 " a list with no extra information, it just finds all possible \+ arrangements of the objects in the list in a similar manner to the pro cedure " }{TEXT 0 7 "arrange" }{TEXT -1 31 " given in the previous sec tion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "permute([A,B,C,D]);" }}{PARA 12 "" 1 "" {XPPMATH 20 " 6#7:7&%\"AG%\"BG%\"CG%\"DG7&F%F&F(F'7&F%F'F&F(7&F%F'F(F&7&F%F(F&F'7&F% F(F'F&7&F&F%F'F(7&F&F%F(F'7&F&F'F%F(7&F&F'F(F%7&F&F(F%F'7&F&F(F'F%7&F' F%F&F(7&F'F%F(F&7&F'F&F%F(7&F'F&F(F%7&F'F(F%F&7&F'F(F&F%7&F(F%F&F'7&F( F%F'F&7&F(F&F%F'7&F(F&F'F%7&F(F'F%F&7&F(F'F&F%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "If you give a number r as a second input parameter, then the procedure " }{TEXT 0 7 "permute" } {TEXT -1 63 " finds the permutations of r objects taken from the given list." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "permute([A,B,C,D],2);\nnops(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7.7$%\"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$F*F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"# 7" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "In this example there are 4 choices for the 1st object and 3 choices for the 2nd object, making a total of 4*3 = 12 arrangements or permutatio ns." }}{PARA 0 "" 0 "" {TEXT -1 87 "As another example, the number of \+ permutations of 3 objects taken from 5 is 5*4*3 = 60." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "permute([ A,B,C,D,E],3);\nnops(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7hn7%%\"AG %\"BG%\"CG7%F%F&%\"DG7%F%F&%\"EG7%F%F'F&7%F%F'F)7%F%F'F+7%F%F)F&7%F%F) F'7%F%F)F+7%F%F+F&7%F%F+F'7%F%F+F)7%F&F%F'7%F&F%F)7%F&F%F+7%F&F'F%7%F& F'F)7%F&F'F+7%F&F)F%7%F&F)F'7%F&F)F+7%F&F+F%7%F&F+F'7%F&F+F)7%F'F%F&7% F'F%F)7%F'F%F+7%F'F&F%7%F'F&F)7%F'F&F+7%F'F)F%7%F'F)F&7%F'F)F+7%F'F+F% 7%F'F+F&7%F'F+F)7%F)F%F&7%F)F%F'7%F)F%F+7%F)F&F%7%F)F&F'7%F)F&F+7%F)F' F%7%F)F'F&7%F)F'F+7%F)F+F%7%F)F+F&7%F)F+F'7%F+F%F&7%F+F%F'7%F+F%F)7%F+ F&F%7%F+F&F'7%F+F&F)7%F+F'F%7%F+F'F&7%F+F'F)7%F+F)F%7%F+F)F&7%F+F)F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#g" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 16 "In general, the " }{TEXT 259 48 "nu mber of permutations of r objects taken from n" }{TEXT -1 3 " is" }} {PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "n*`.`*(n-1)*`.`*(n-2 )*` . . . `*(n-r+1)" "6#*0%\"nG\"\"\"%\".GF%,&F$F%F%!\"\"F%F&F%,&F$F% \"\"#F(F%%(~.~.~.~GF%,(F$F%%\"rGF(F%F%F%" }{TEXT -1 2 ". " }}{PARA 0 " " 0 "" {TEXT -1 57 "A convenient expression for this involving factori als is " }{XPPEDIT 18 0 "n!/(n-r)!;" "6#*&-%*factorialG6#%\"nG\"\"\"-F %6#,&F'F(%\"rG!\"\"F-" }{TEXT -1 58 ", and this expression is denoted by the symbol P(n,r) or " }{XPPEDIT 18 0 "``[n];" "6#&%!G6#%\"nG" } {XPPEDIT 18 0 "P[r];" "6#&%\"PG6#%\"rG" }{TEXT -1 1 "." }}{PARA 258 " " 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 1 " " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{TEXT 271 10 "__________" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "For example, the number of arrangements of 5 objects take n 3 at a time is" }}{PARA 256 "" 0 "" {TEXT -1 10 " P(5,3) = " } {XPPEDIT 18 0 "5!/2! = 5*`.`*4*`.`*3*`.`*2*`.`*1/(2*`.`*1);" "6#/*&-%* factorialG6#\"\"&\"\"\"-F&6#\"\"#!\"\"*6F(F)%\".GF)\"\"%F)F/F)\"\"$F)F /F)F,F)F/F)F)F)*(F,F)F/F)F)F)F-" }{TEXT -1 6 " = 60." }}{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 "combi nat" }{TEXT -1 27 " performs this calculation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "numbperm(5,3 );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#g" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "The number of arrangements of 2 6 letters taken 3 at a time is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "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 26 "Selections or combinations" }}{PARA 0 "" 0 "" {TEXT 267 8 "Question" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 71 "In ho w many ways can we choose 3 people from 7 to serve on a committee?" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 8 "Solution " }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 73 "We start by obtainin g all possible arrangements of 3 people taken from 7." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "with(comb inat):\nperm := permute([A,B,C,D,E,F,G],3);\nnops(perm);" }}{PARA 7 " " 1 "" {TEXT -1 67 "Warning, the protected name Chi has been redefined and unprotected\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%permG7^x7%%\" AG%\"BG%\"CG7%F'F(%\"DG7%F'F(%\"EG7%F'F(%\"FG7%F'F(%\"GG7%F'F)F(7%F'F) F+7%F'F)F-7%F'F)F/7%F'F)F17%F'F+F(7%F'F+F)7%F'F+F-7%F'F+F/7%F'F+F17%F' F-F(7%F'F-F)7%F'F-F+7%F'F-F/7%F'F-F17%F'F/F(7%F'F/F)7%F'F/F+7%F'F/F-7% F'F/F17%F'F1F(7%F'F1F)7%F'F1F+7%F'F1F-7%F'F1F/7%F(F'F)7%F(F'F+7%F(F'F- 7%F(F'F/7%F(F'F17%F(F)F'7%F(F)F+7%F(F)F-7%F(F)F/7%F(F)F17%F(F+F'7%F(F+ F)7%F(F+F-7%F(F+F/7%F(F+F17%F(F-F'7%F(F-F)7%F(F-F+7%F(F-F/7%F(F-F17%F( F/F'7%F(F/F)7%F(F/F+7%F(F/F-7%F(F/F17%F(F1F'7%F(F1F)7%F(F1F+7%F(F1F-7% F(F1F/7%F)F'F(7%F)F'F+7%F)F'F-7%F)F'F/7%F)F'F17%F)F(F'7%F)F(F+7%F)F(F- 7%F)F(F/7%F)F(F17%F)F+F'7%F)F+F(7%F)F+F-7%F)F+F/7%F)F+F17%F)F-F'7%F)F- F(7%F)F-F+7%F)F-F/7%F)F-F17%F)F/F'7%F)F/F(7%F)F/F+7%F)F/F-7%F)F/F17%F) F1F'7%F)F1F(7%F)F1F+7%F)F1F-7%F)F1F/7%F+F'F(7%F+F'F)7%F+F'F-7%F+F'F/7% F+F'F17%F+F(F'7%F+F(F)7%F+F(F-7%F+F(F/7%F+F(F17%F+F)F'7%F+F)F(7%F+F)F- 7%F+F)F/7%F+F)F17%F+F-F'7%F+F-F(7%F+F-F)7%F+F-F/7%F+F-F17%F+F/F'7%F+F/ F(7%F+F/F)7%F+F/F-7%F+F/F17%F+F1F'7%F+F1F(7%F+F1F)7%F+F1F-7%F+F1F/7%F- F'F(7%F-F'F)7%F-F'F+7%F-F'F/7%F-F'F17%F-F(F'7%F-F(F)7%F-F(F+7%F-F(F/7% F-F(F17%F-F)F'7%F-F)F(7%F-F)F+7%F-F)F/7%F-F)F17%F-F+F'7%F-F+F(7%F-F+F) 7%F-F+F/7%F-F+F17%F-F/F'7%F-F/F(7%F-F/F)7%F-F/F+7%F-F/F17%F-F1F'7%F-F1 F(7%F-F1F)7%F-F1F+7%F-F1F/7%F/F'F(7%F/F'F)7%F/F'F+7%F/F'F-7%F/F'F17%F/ F(F'7%F/F(F)7%F/F(F+7%F/F(F-7%F/F(F17%F/F)F'7%F/F)F(7%F/F)F+7%F/F)F-7% F/F)F17%F/F+F'7%F/F+F(7%F/F+F)7%F/F+F-7%F/F+F17%F/F-F'7%F/F-F(7%F/F-F) 7%F/F-F+7%F/F-F17%F/F1F'7%F/F1F(7%F/F1F)7%F/F1F+7%F/F1F-7%F1F'F(7%F1F' F)7%F1F'F+7%F1F'F-7%F1F'F/7%F1F(F'7%F1F(F)7%F1F(F+7%F1F(F-7%F1F(F/7%F1 F)F'7%F1F)F(7%F1F)F+7%F1F)F-7%F1F)F/7%F1F+F'7%F1F+F(7%F1F+F)7%F1F+F-7% F1F+F/7%F1F-F'7%F1F-F(7%F1F-F)7%F1F-F+7%F1F-F/7%F1F/F'7%F1F/F(7%F1F/F) 7%F1F/F+7%F1F/F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$5#" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "The first arrange ment in this list is [A,B,C]." }}{PARA 0 "" 0 "" {TEXT 259 82 "All arr angements of these 3 objects among themselves are present in the above list" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 37 "We can check vis ually or as follows. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "First find all arrangements of A, B and C." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "sub \+ := permute([A,B,C]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$subG7(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 inters ection 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'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 146 "Since the order is not important for a selection, all these 6 arrange ments really constitute just the one selection of 3 people namely A, B and C." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "The 2nd arrangement among the 210 arrangements of the 7 people tak en 3 at a time is [A,B,D]." }}{PARA 0 "" 0 "" {TEXT -1 112 "Again we f ind that all the arrangements of these 3 people among themselves are i ncluded in the 210 arrangements." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "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 54 " Given a list of arrangements, the following procedure " }{TEXT 0 12 "c ombineperms" }{TEXT -1 144 " will group together all those arrangement s 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 0 621 "combineperms := proc(permut ations::listlist)\n local j,n,groups,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)\} th en\n perms := [op(1..j-1,perms),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 return [groups];\nend proc:" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "perm := permute([A,B,C,D,E,F ,G],3):\ncombineperms(perm);\nnops(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7E7(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'%\"FG7%F&F@F'7%F'F&F@7%F'F@F& 7%F@F&F'7%F@F'F&7(7%F&F'%\"GG7%F&FHF'7%F'F&FH7%F'FHF&7%FHF&F'7%FHF'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&F(F@7%F&F@F(7%F(F&F@7%F(F@F&7%F@F&F (7%F@F(F&7(7%F&F(FH7%F&FHF(7%F(F&FH7%F(FHF&7%FHF&F(7%FHF(F&7(7%F&F0F87 %F&F8F07%F0F&F87%F0F8F&7%F8F&F07%F8F0F&7(7%F&F0F@7%F&F@F07%F0F&F@7%F0F @F&7%F@F&F07%F@F0F&7(7%F&F0FH7%F&FHF07%F0F&FH7%F0FHF&7%FHF&F07%FHF0F&7 (7%F&F8F@7%F&F@F87%F8F&F@7%F8F@F&7%F@F&F87%F@F8F&7(7%F&F8FH7%F&FHF87%F 8F&FH7%F8FHF&7%FHF&F87%FHF8F&7(7%F&F@FH7%F&FHF@7%F@F&FH7%F@FHF&7%FHF&F @7%FHF@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'F(F@7%F'F@F(7%F(F'F@7%F(F @F'7%F@F'F(7%F@F(F'7(7%F'F(FH7%F'FHF(7%F(F'FH7%F(FHF'7%FHF'F(7%FHF(F'7 (7%F'F0F87%F'F8F07%F0F'F87%F0F8F'7%F8F'F07%F8F0F'7(7%F'F0F@7%F'F@F07%F 0F'F@7%F0F@F'7%F@F'F07%F@F0F'7(7%F'F0FH7%F'FHF07%F0F'FH7%F0FHF'7%FHF'F 07%FHF0F'7(7%F'F8F@7%F'F@F87%F8F'F@7%F8F@F'7%F@F'F87%F@F8F'7(7%F'F8FH7 %F'FHF87%F8F'FH7%F8FHF'7%FHF'F87%FHF8F'7(7%F'F@FH7%F'FHF@7%F@F'FH7%F@F HF'7%FHF'F@7%FHF@F'7(7%F(F0F87%F(F8F07%F0F(F87%F0F8F(7%F8F(F07%F8F0F(7 (7%F(F0F@7%F(F@F07%F0F(F@7%F0F@F(7%F@F(F07%F@F0F(7(7%F(F0FH7%F(FHF07%F 0F(FH7%F0FHF(7%FHF(F07%FHF0F(7(7%F(F8F@7%F(F@F87%F8F(F@7%F8F@F(7%F@F(F 87%F@F8F(7(7%F(F8FH7%F(FHF87%F8F(FH7%F8FHF(7%FHF(F87%FHF8F(7(7%F(F@FH7 %F(FHF@7%F@F(FH7%F@FHF(7%FHF(F@7%FHF@F(7(7%F0F8F@7%F0F@F87%F8F0F@7%F8F @F07%F@F0F87%F@F8F07(7%F0F8FH7%F0FHF87%F8F0FH7%F8FHF07%FHF0F87%FHF8F07 (7%F0F@FH7%F0FHF@7%F@F0FH7%F@FHF07%FHF0F@7%FHF@F07(7%F8F@FH7%F8FHF@7%F @F8FH7%F@FHF87%FHF8F@7%FHF@F8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#N " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 199 "The 210 arrangements of 7 people taken 3 at a time can be grouped togethe r in 35 groups of 6 where each group of 6 arrangements just involves a rranging a given selection of 3 people amomg themselves." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 "Each of the 35 gr oups of arrangements gives rise to just one selection in which we igno re the order of the 3 people." }}{PARA 0 "" 0 "" {TEXT -1 64 "The numb er of ways of choosing 3 people from 7 is therefore 35." }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{TEXT 269 11 "___________" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "A selecti on of objects in no particular order is called a " }{TEXT 259 11 "comb ination" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 122 "Since the ord er of the objects in a selection or combination is not important, we c an represent 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 repr esent 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 19 "The next procedure " }{TEXT 0 12 "permstocombs" }{TEXT -1 163 " is similar to the previous procedure except that each of the \+ groups is replaced by a single set to represent the associated combina tion. 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 0 529 "permstocombs := proc(permut ations::listlist)\n local j,n,combs,p,perms;\n perms := permutatio ns;\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 per ms := [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 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Applying this procedure to the var iable perm" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 64 "perm := permute([A,B,C,D,E,F,G],3):\npermstocombs(p erm);\nnops(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7E<%%\"CG%\"AG%\"BG <%%\"DGF&F'<%%\"EGF&F'<%%\"FGF&F'<%%\"GGF&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 52 "This shows all the ways o f choosing 3 people from 7." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 274 "If we are just interested in calculating the n umber of ways of selecting 3 people from 7, we can obtain this number \+ by dividing the number of arrangements of 3 people from 7 by the numbe r of ways of arranging 3 people in each selection of 3 amomng themselv es, namely 3! = 6." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "Thus the number of ways of choosing 3 things from 7 is \+ " }{XPPEDIT 18 0 "P(n,r)/3!;" "6#*&-%\"PG6$%\"nG%\"rG\"\"\"-%*factoria lG6#\"\"$!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "7!/(4!*3!);" "6#*&-% *factorialG6#\"\"(\"\"\"*&-F%6#\"\"%F(-F%6#\"\"$F(!\"\"" }{TEXT -1 1 " ." }}{PARA 0 "" 0 "" {TEXT -1 16 "In general, the " }{TEXT 259 51 "num ber of ways of choosing r objects from n objects" }{TEXT -1 41 " (with out regard to order) is denoted by " }{XPPEDIT 18 0 "C(n,r)" "6#-%\"CG 6$%\"nG%\"rG" }{TEXT -1 4 " or " }{XPPEDIT 18 0 "``[n];" "6#&%!G6#%\"n G" }{XPPEDIT 18 0 "C[r];" "6#&%\"CG6#%\"rG" }{TEXT -1 1 "." }}{PARA 256 "" 0 "" {TEXT -1 3 " " }{XPPEDIT 18 0 "C(n,r)" "6#-%\"CG6$%\"nG% \"rG" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "P(n,r)/r!;" "6#*&-%\"PG6$%\"n G%\"rG\"\"\"-%*factorialG6#F(!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " n!/((n-r)!*r!);" "6#*&-%*factorialG6#%\"nG\"\"\"*&-F%6#,&F'F(%\"rG!\" \"F(-F%6#F-F(F." }{TEXT -1 1 "." }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {TEXT 270 13 "_____________" }{TEXT -1 1 " " }}{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 27 " performs this calculation." }}{PARA 0 "" 0 "" {TEXT -1 51 "The number of ways of selecting 3 people from 7 is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "with(co mbinat):\nnumbcomb(7,3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#N" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "For anoth er example, consider the selecting 3 people from the 4 people A, B, C \+ and D." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "The number of ways of arranging 4 objects 3 at a time is 24." }} {PARA 0 "" 0 "" {TEXT -1 156 "On the other hand, the number of ways of choosing 3 objects from 4 is just 4, since there are 4 possibilities \+ for an object to leave out from the selection." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "with(combin at):\nperm := permute([A,B,C,D],3);\nnops(perm);\ncombineperms(perm); \npermstocombs(perm);\nnops(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%% permG7:7%%\"AG%\"BG%\"CG7%F'F(%\"DG7%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#\"#C" }}{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(F07 %F&F0F(7%F(F&F07%F(F0F&7%F0F&F(7%F0F(F&7(7%F'F(F07%F'F0F(7%F(F'F07%F(F 0F'7%F0F'F(7%F0F(F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&<%%\"CG%\"AG% \"BG<%%\"DGF&F'<%F)F%F&<%F)F%F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\" \"%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "numbcomb(4,3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\" \"%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "T he Maple procedure " }{TEXT 0 6 "choose" }{TEXT -1 16 " in the package " }{TEXT 0 8 "combinat" }{TEXT -1 37 " can be used to enumerate selec tions." }}{PARA 0 "" 0 "" {TEXT -1 85 "The original objects 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 29 "choose( \{A,B,C,D\},3);\nnops(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<&<%%\"DG %\"CG%\"BG<%F%F&%\"AG<%F&F)F'<%F%F)F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "choose(\{A,B,C,D,E,F,G\},3);\nnops(%);" }}{PARA 12 " " 1 "" {XPPMATH 20 "6# " 0 "" {MPLTEXT 1 0 50 "choose(\{A,B,C,D,E,F,G\},4);\nnops(%);\nnumbcomb(7,4) ;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6# " 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 -1 176 "Find the number of different flags which can be made whi ch consist of three different vertical coloured bands where the possib le colours are red, green, blue, yellow and black." }}{PARA 0 "" 0 "" {TEXT -1 66 "Make a list of the permutations which describe the possib le flags." }}{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 -1 198 "Find the number of different flags which can be made which consist of three different vertical coloured bands \+ where the possible colours are red, green, blue, cyan, yellow, magenta , black and white." }}{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 87 "(a) Find the number of ways of choos ing 2 people from the 6 people A, B, C, D, E and F." }}{PARA 0 "" 0 " " {TEXT -1 34 " List the possible selections." }}{PARA 0 "" 0 "" {TEXT -1 87 "(b) Find the number of ways of choosing 4 people from the 6 people A, B, C, D, E and F." }}{PARA 0 "" 0 "" {TEXT -1 34 " Li st the possible selections." }}{PARA 0 "" 0 "" {TEXT -1 58 "(c) Why ar e the numerical answers to (a) and (b) the same?" }}{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 "Q4 " }} {PARA 0 "" 0 "" {TEXT -1 81 "(a) Find the number of ways of choosing 2 cards from a standard deck of 52 cards." }}{PARA 0 "" 0 "" {TEXT -1 81 "(b) Find the number of ways of choosing 5 cards from a standard de ck of 52 cards." }}{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 ";" }}}}{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 17 "Code for pictures" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 21 "Code for tree diagram" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 663 "with(plots) :\nlines := PLOT(CURVES([[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]],\n[[1,-3],[2,-2.5]],[[1,-1],[2,-1.5]],[[1, -1],[2,-0.5]],\nCOLOR(RGB,0,0,0)),\nPOINTS([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],SYMBOL(CIRCLE))):\ntext := textplot([[.8,1.9,`gold`],[.7,.5,`white` ],[.7,-.3,`green`],\n[.9,-2,`maroon`],[1.5,3.6,`manual`],[1.5,2.5,`aut omatic`],\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`]]): \ndisplay(\{lines,text\},axes=none);" }}}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "4 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }