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.
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