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CourseworkHelp
: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
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/2
formular 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 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.
Use this formular, we can find out the total arrangements of all numbers and letters.
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