Emmas Dilemma Essay Research Paper Firstly we

Emma?s Dilemma Essay, Research Paper Firstly we arrange EMMA?s Name. 1)EAMM 7)MAEM 2)EMAM 8)MAME 3)MEMA 9)AMME 4)MEAM 10)AEMM 5)MMEA 11)AMEM 6)MMAE 12)EMMA Secondlywe arrange lucy?s name. 1)Lucy 12)Cyul 22)Yulc 2)Luyc 13)Culy 23)Ycul 4)Lycu 14)Culy 24)Yluc 5)Lcuy 15)Cylu 25)Ucyl 6)Lcyu 16)Clyu 7)Ulcy 17)Cuyl 8)Ucly 18)Yluc 9)Uycl 19)Yucl 10)Ulyc 20)Yclu 11)Uylc 21)Ylcu From these 2 investigation I worked out a method: Step1: 1234—Do the last two number first then you get 1243. 1243—Do the last three numbers and try the possibility. 1423. 1432. 1342. 1324, because the number 2 has been the first number of last three numbers, so we don?t do it again. Step2: we have list all arrangements of 1 go front, so we do 2 go front. 2134 and we do same thing to it, it will like this: 2134—2143, 2143—2431,2413,2314,2341 Step3: We have finished 2 go first, then let?s do 3 go ahead. 3124—3142, 3142—3241,3214,3412,3421 Step4: We have finished 3 go ahead, then try 4 4123—4132, 4132—4231,4213,4312,4321 We have list all arrangement of 1234, use this method we can arrange the number which has 5 figures or more. We are trying to work out a formula which can calculate the number of arrangement when we look at a number. Let?s list all the arrangment for 1234: 1234 4123 1243 4132 1324 — 6 arrangment 4231 —- 6 arrangement 1342 4213 1432 4321 1423 4312 2134 3124 2143 3143 2341 —- 6 arrangements 3241 — 6 arrangements 2314 3214 2413 3412 2431 3421 So if we time 6 by 4, we would get 24, and we can get the total arrangements of 24. Let?s try 5 figures: 12345 13245 12354 13254 12435 — 6 arrangements 13452 — 6 arrangements 12453 13425 12534 13524 12543 13542 14235 15432 14253 15423 14352 — 6 arrangements 15324 —- 6 arrangements 14523 15342 14532 15243 14325 15234 Don?t you notice the arrangements of last 4 numbers added up, it equal 24, so if we time 24 by 5, and get 120, and 120 is the total of arrangements. Carry on, if a number has 6 figure, then the total of arrangement should be 120 times by 6, and get 720, and 720 is the total of arrangements. Carry on, if a number has 7 fugure, then the total of arrangement should be 720 times by 7, and get 5040, the total of arrangement is 5040. This is my prediction, let?s work it out a formula, and confirm it. 3 figure with different number it has 6 arrangements 4 4*6 5 4*6*5 6 4*6*5*6 7 4*6*5*6*7 8 4*6*5*6*7*8 so on We can rewrite it as: 1 fig 1 2 fig 1*2 3 fig 1*2*3 4 fig 1*2*3*4 5 fig 1*2*3*4*5 6 fig 1*2*3*4*5*6 so on There?s a simbol for the frequancy above, that?s I.For example: 1*2 = 2i 1*2*3 = 3i 1*2*3*4 = 4i so on So if n represent the number of figures of a number, then it has arrangements of ni. The formula: NI NI: Can be caculated on caculator. Process: pres key N (the number of figure), then press key I, then you would get the arrangements. Let?s confirm this formula, if a nuber has 1fig it has 1 arrangements formular: 1*1=1 It works 2fig it has 2 arrangements formular: 1*2=2 It works 3fig it has 6 arrangements formular: 1*2*3=6 It works 4fig it has 24 arrangements formular: 1*2*3*4=24 It works Formula is confirmed What about if a number has two same figure For example: 223, 334 Let?s try to work out the frequency of them 223 can be arranged as 232, 322 only 3 arrangements try 4 figures with 2 same numbers 1223 arranged as: 1223 2123 3122 1232 —- 3 arrangements 2132 3212 — 3 arrrangements 1322 2213 —- 6 arrangements 3221 2231 2312 2321 total arrangement is 12 Try 5 fig: 42213 12234 42231 12243 42123 12324 42132 12342 42321 12423 42312 ——– 12 arrangements 12432 ——– 12 arrangements 41223 13224 41232 13242 41322 13422 43122 14223 43212 14232 43221 14322 21234 23124 31224 21243 23142 31242 21324 23214 31422 21342 23241 32124 21423 23421 32142 21421 23412 —–24 arrangments 32214 ——- 12 arrangements 22134 24123 32241 22143 24132 32412 22314 24231 32421 22341 24213 34122 22413 24312 34212 22431 24321 34221 so the total arrangements are 12*5=60 We have found the frequency 2 figure with 2 same number 1arrangements 3 1*3 4 1*3*4 5 1*3*4*5 Let?s work out the formular: if n= number of figures a= number of arrangements the formular is a=ni/2 Let?s confirm the formular: 2 fig with 2 same number formular: 2/2=1 it works 3 (1*2*3)/2=3 it works 4 (1*2*3*4)/2=12 it works Formular is confirmed What about if 3 numbers are the same let?s try 333 only on arrangement Try 3331 3331 3313 —— 4 arrangements 3133 1333 Try 33312 33312 31233 12333 33321 31323 13233 —4 arrangements 33123 31332 —-12 arrangements 13323 33132 32331 13332 33231 32313 33213 32133 21333 23133—-4 arrangements 23313 23331 Total arrangements are 4*5=20 Let?s try 6 fig with 3 same number 333124 332134 333142 332143 333214 334321 333241 332314 333412 332341 —–24 arrangements 333421 332413 331234 332431 331243 334123 331324 334132 331342 334213 331423 334231 331432 334312 312334 321334 312343 so on —-12 arrangements 312433 313234 313243 34——- 313324 — 12 arrangements so on —–12 arrangements 313342 313423 313432 314233 314323 314332 123334 133324 2—- 123343 133342 so on ——–20 arrangements 123433 133234 124333 133243 124332 133423 — 20 arrangements 4—- 134323 133432 so on ———20 arrangements 134233 142333 132334 143233 132343 143323 132433 143332 Total arrangement for 6 figure with 3 same number is 120, 20*6 Let?s see the construction: 3 fig with 3same number 1 arrangement 4 1*4 5 1*4*5 6 1*4*5*6 Can you see the pattern?so the formula for three sames numbers of a number is: a= ni/6 let?s review the formula: formula for different number: a=ni formula for 2 same number: a=ni/2formular for 3 same number: a=ni/6 Let?s put them is this way: n 1 2 3 x 1 2 6 n represent the number of figures of a number x represent the divided number in the formular Do you notice that x equal the last x time n, so I expect the formula for 4 sames number of a number is: a= ni/24 Let?s confirm it: try 4 same number. 4 fig, one arrangement. a=n/24=(1*2*3*4)/24=1 the formular works try 5 figures 11112 11121 11211 —– 5 arrangements 12111 21111 a=n/24=(1*2*3*4*5)/24=5 the formular works So formula is confirmed Let?s investigate the formula, and improve it. n 1 2 3 4 5 x 1 2 6 24 110 so the formula for this is x=ni so if A represent arrangement, and n represent numbers of figures, x represent the number fo same number, and the formula is: a=ni/xi *notic I can not be cancel out. What about if a number has 2 pairs of same number. what would happen to the formula. Let?s try 4 fig with 2 pairs of 2 same number. 1122 2122 1212 2212 —– 6 arrangements 1221 2221 let?s see if the formula still work a=(1*2*3*4)/2=12 No, it doesn?t work but if we divide it by two. let?s try 6 fig, with 2 pairs of 3 same numbers 111222 121212 222111 211212 112122 121122 221211 211221 112212 122112 –10 arrangements 221121 212112 —10 arrangements 112221 122121 221112 212121 121221 122211 211122 212211 total arrangement is 20 let?s see the formular: a=(1*2*3*4*5*6)/1*2*3 =120 no, it doesn?t work but if we divide it by another 6 which is (1*2*3) Can you see the pattern, the formular still work if we times the mutiply again. For example: for 4 figures with 2 pairs of 2 same number. a=(1*2*3*4)/(1*2*1*2)=6 it works so I expect it still work for 6 figures with 2 pairs of 3 same number. follow this formula, I predict the arrange for this is a=(1*2*3*4*5*6)/1*2*3*1*2*3=20 111222 121221 122121 2—– 112122 121212 122211 ——– 10 arrangement so on ——10 arrangements 112212 121122 112221 122112 Total arrangement= 20 the formula work I expect the arrangements for 8 fig will be a= (1*2*3*4*5*6*7*8)/1*2*3*4*1*2*3*4=70 let?s confirm 11112222 11211222 11221212 11121222 11212122 11221221 11122122 11212212 11222112 ——— 15 arrangements 11122212 11212221 11222121 11122221 11221122 11222211 12111222 12122112 12212121 12121212 12112122 12122121 12212211 12211212 12112212 12122211 12221112 ————–20 arrangements 12112221 12211122 12221121 12121122 12211221 12221211 12121221 12212112 12222111 2——- so on ——35 arrangement Total arrangement is 70, it works the formular is confirmed This formula can be written as: a=ni/xixi x represent the number of figures of same number What about a number with difference number of figure of same number. For example: 11122,111122 let?s try if the formula still work. The formula is a=ni/xixi but we need to change the formula, because there are 2 pairs of same numbers with different number of figures. so we change the formula to a=ni/x1i*x2i Let?s try 5 figures with 3 same number, and 2 same number. According to the formula, I expect the total arrangement for this is a=(1*2*3*4*5)/(3*2*1*2*1)=10 11122 12211 11212 21112 11221 21121 ——-10 arrangements 12112 21211 12121 22111 The formular still works. Let?s try 7 fig, with 3 same number, and 4 same number. I expect the total arrangement is a=(1*2*3*4*5*6*7)/(3*2*1*4*3*2*1)=35 1112222 1222211 2222111 2211212 1121222 1222121 2221211 2211221 1122122 1222112 2221121 2212112—–10 arrangements 1122212 1221221—–15 arrangements 2221112 2212121 1122221 1221212 2211122 2212211 1212221 1221122 1212212 1212122 1211222 2111222 2112221 2121221 2122211 2112122 2121122 2122112 ——— 10 arrangements 2112212 2121212 2122121 Total arrangment is 35, the formular works. The formular is confirmed. What about three pairs of same number The formmular need to be rewritten as a=ni/xixixi There are three xi need to mutiple ni, because there are three pairs of same number. if there are two pair of same number of figures of same number, then there are only two xi need to mutiply, and if there are two pair of different number of figures of same number, then there would be x1i and x2i need to mutiply. Let?s confirm the formular. let?s try 112233, according the formula, I expect the total arrangements is a=(1*2*3*4*5*6)/(1*2*1*2*1*2)=90 Let?s confirmed 112233 121233 123123 131223 132231 112323 121323 123132 131232 132123 112332 121332 123213 131322 132132 —— 30 arrangements 113223 122133 123231 132321 133122 113232 122313 123312 132312 133212 113322 122331 123321 132213 1332212——- 3—— so on ——30 arrangements so on ——— 30arrangements The total arrangement is 90, the formular works. Formular is confirmed What about three pairs of different number of figures of a number For example: 122333 according the formula, the total arrangment is a=(1*2*3*4*5*6)/(1*1*2*1*2*3)=60 Let?s confirm it: 122333 212333 231332 3——– 123233 213233 232133 so on ——- 30 arrangements 123323 213323 232313 123332 213332 232331 —–30 arrangements 132233 221333 233123 132323 223133 233132 132332 223313 133213 133223 223331 233231 133232 231233 233312 133322 231323 233321 The formular works Formular is confirmed From the investigation above we find out the formular for calculating the number of arrangements, it?s a=ni/xi a represent the total arrangements n represent the number of figures of the number I represent the key I x represent the numbers of figures of same number of the number if there are more than one pair of same number, x2, or x3, so on may added to the formular, it depend how many pairs of same number. For example: for 2 pairs of same number of figures of same number of a number the formula is a=ni/xixi for 2 pairs of different number of figures of same number of a number the formula is a=ni/x1ix2i for 3 pairs of same number of figures of same number of a number the formula is a=ni/xixixi form 3 pairs of different number of figures of same number of a number the formular is a=ni/x1ix2ix3i. The formular can be also used to the arrangements of letter. For example: xxyy the arrangement for this is a=(4*3*2*1)/(2*1*2*1)=6 xxyyy the arrangement for this is a=(5*4*3*2*1)/(3*2*1*2*1)=10 xxxxxxyyyyyyyyyy the arrangement for this is a=ni/x1ix2i=(16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1)/(10*9*8*7*6*5*4*3*3*2*1*6*5*4*3*2*1)=8008 The total arrangement is 8008. Use this formular, we can find out the total arrangements of all numbers and letters.