• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Emma's Dilemma

Extracts from this document...

Introduction

                                                                                                                Ehsan Tajalifar

image00.png

image01.png

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

                                                       AMME

                                                       AMEM

                                                       AEMM

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.

LUCY      

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.

Middle

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.

2) JACK

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

ET                                          

TE

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.

Conclusion

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)

AS                                                        

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) !        

NA                                        

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,

5!  

A and mentioned above, this gives  5P2  =        

3!      

                                                                   =     20 as requirements

we know that there are n! arrangements of n objects  

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

              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)!

image08.png

n!     m!

...read more.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Emma's Dilemma

    of = No. of letters ! combinations ( times a letter X ( times a different letter has been repeated ) has been repeated ) By writing this, you still mean the previous line, but it is now a lot easier to write down, especially if the number of letters exceeds 10.

  2. Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

    I learned this formula form background research I did myself, and have decided to incorporate it in to my work. Factorials can be used, by instead of writing: "No. of letters X ( No. of letters - 1 ) X ( No.

  1. I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

    number of arrangements: AABC: Earlier in my investigation I worked out that for a word with four letters in it, two of which are the same, there are 12 different combinations. As listed below: AABC AACB ABCA ABAC ACAB ACBA BAAC BACA BCAA CAAB CABA CBAA Using my formula I can also check the number of different permutations.

  2. Psychologists have identified several "laws" of perceptual organisation on grouping which illustrate their view ...

    Common fate: Elements seen moving together are perceived as belonging together. This is why a group of people running in the same direction appear to be unified in their purpose. Navons 1977 experimental test of Gestalt laws Navon tested the idea that the whole is perceived before the parts that

  1. To investigate the combination of arrangement of letters in Jeans name and then for ...

    E S M J A E M S J M A E S J M A S E J M S E A J M S A E J M E S A J M E A S J E S A M J E S M A J E

  2. Emma’s Dilemma.

    Altogether I have come up with 120 permutations for a name with five different letters. Judging from the results I have got so far, if I continue with even longer names then there are going to be too many permutations to document.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work