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  • Level: GCSE
  • Subject: Maths
  • Word count: 2360
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    EMMA's Dilemma Emma and Lucy

    Extracts from this document...

    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

    ...read more.

    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:

    ...read more.

    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.

    ...read more.

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