EMMA's Dilemma Emma and Lucy

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 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.

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.