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.