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EMMA's Dilemma Emma and Lucy
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Introduction
EMMA’s Dilemma
We are investigating the number of different arrangements of letters.
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
Middle
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:
Conclusion
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 133221
2------- 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.
This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.
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