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Emma's Dilemma

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                                                                                                                Ehsan Tajalifar



1) Investigate the number of the different arrangements of the letters of EMMA ‘s name.

   EMMA*                                     There is 18 different arrangements

   EMAM                                       of the letters of EMMA, but 6 of                                                    

   EAMM                                       them are the same so there are 12

   MEMA                                       different arrangements of the letter

   MEAM                                       EMMA.

   MAEM                                       In EMMA, M is duplicated  

   MAME                                       because of this we have 12                      

   MMAE                                       duplicated arrangements in our    

   MMEA                                       list.

  AMME                                        EMMA  

  AMEM                                        EMAM

  AEMM                                        EAMM  

  EMMA*                                      MAEM    

  EAMM                                        MAME

  MMAE                                        MMEA

  AEMM                                        MMAE

  MMEA                                        MEAM                        

  AMME                                        MEMA




The other thing that I found out was that in JACK there are 24 different arrangements but in EMMA there are only 12 different arrangements even though they both have 4 letters. This is because in EMMA, M is duplicated and because of this some of these arrangements that we get are the same so we don’t count them.                                                                                                                    

2) Investigate the number of the different arrangements of the letter s of LUCY ‘s name.


LCUY                UCLY                CLUY                YCLU        image06.png

LCYU                UCYL                CLYU                YCUL

LUCY                ULCY                CULY                YLCU

LUYC                ULYC                CUYL                YLUC

LYCU                UYLC                CYLU                YUCL

LYUC                UYCL                CYUL                YULC

There are 6 different arrangements for each letter. And we got four different letters by multiplying 4 by 6 we get 24, which is the number of different arrangements of the letters of LUCY’s Names.

...read more.


SAHEN        SENAH        SHNAE        SANEH          letter S.  

SAHNE        SENHA        SHNEA        SANHE

AEHNS        AHENS        ANEHS        ASEHNimage09.png

AEHSN        AHESN        ANESH        ASENH        6x4=24  

AENHS        AHNES        ANHES        ASHEN         There is 24 different                                                                                                                      

AENSH        AHNSE        ANHSE        ASHNE        arrangements for

AESHN        AHSEN        ANESH        ASNEH        letter A.

AESNH         AHSNE        ANEHS        ASNHE

NAEHS        NEAHS        NHAES        NSAEH          6x4=24image09.png

NAESH        NEASH        NHASE        NSAHE        

NAHES        NEHAS        NHEAS        NSEAH          There is 24 different                                                                                                                                                          NAHSE         NEHSA        NHESA        NSEHA          arrangements for

NASEH        NESAH        NHSAE        NSHAE          letter N.

NASHE        NESHA        NHSEA        NSHEA          

When we add all of them up we get 120, which is the total combination of the letter EHSAN.


JACK                ACJK                   CAJK                KACJ        

JAKC                ACKJ                CAKJ                KAJC              

JCAK                AKCJ                CJKA                KCAJ              

JCKA                AKJC                CJAK                KCJA              

JKAC                AJCK                CKAJ                KJAC              

JKCA                AJKC                CKJA                KJCA

Each letter has 6 different arrangements and we have 4 different letters, in order to find out how many different Arrangements we have to multiply 4

by 6,which is 24.

3) ALI                                                                                                                                                                                          

AIL                IAL                LAI                                          

ALI                ILA                LIA        

Each letter has 2 different arrangements and we have 3 different letters and in order to find out different Arrangements we have to multiply 2 by 3,which is 6.

4) ET



Each letter has one different arrangement and we have 2 different letters and in order to find out different arrangements we have to multiply 1 by two, which, is 2.

...read more.


EN                                        ways the second object can be chosen

NE                                        in (n - 1) ways, the third object can be

HS                                        chosen in (n - 2) ways… and so on until    

SH                                        the 4th object which can be written in

HA                        20                (n – r + 1) ways.

AH                                        So the formula to find permutations

HN                                        of r objects chosen from n unlike objects

NH                        is:

SA                                              n (n - 1) (n – 2) (n - 3) (n – 4) …. (n - r + 1)


SN                                              n (n - 1) (n – 2) … (n - r + 1)            n !

NS                                             =                                                 =  image04.pngimage05.png

AN                  (n – r)(n – r – 1) … 2 x 1           (n - r) !        


We usually say that the number of permutations of r objects selected from

                                                                 n !

 n unlike objects is “P r where P r =      

                                                              (n - r) !        

Thus in the case of arranging two letters chosen from the five letter E,H,S,


A and mentioned above, this gives  5P2  =        


                                                                   =     20 as requirements

we know that there are n! arrangements of n objects  

chosen from n objects( is n! ). Thus “P,, must equal n !  


i.e.                         = n!

          (n - n) !        

and we therefore define 0! as 1.

4) XX …XXYY...YY

Investigate the number of different arrangements of the letters.

We assume we have n number of X and m number of Y

XX…XXX     YY…YYYimage07.png

         n                    m

According to my pervious investigation I can write the number of different arrangements of X ‘s and Y ‘ as below.  

(n + m)!


n!     m!

...read more.

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