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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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