• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  • Level: GCSE
  • Subject: Maths
  • Word count: 1116

Emma’s Dilemma

Extracts from this document...

Introduction

Emma’s Dilemma

In my investigation I am going to investigate of the different arrangements of letters in a word. My aim is to be able to generalise a formula  for all my findings.

E.g. Tim           Is one arrangement

        Mit            Is another

  • First I am going to investigate how many different arrangements in the name “Emma”, which has no letters the same.

EMMA         AMME            AMEM       EMAM         AEMM           EAMM             MMEA    

MMAE         MEMA            MAME      MEAM       MAEM

 4-letter word, 2 letters repeated, 12 different arrangements.  

Now I am going to investigate the different arrangement of Emma’s friends name “Lucy”

LUCY         LCUY          LYUC     LUYC        LCYU     LYCU

 UCLY          UYCL              UYLC     ULYC        UCYL     ULCY 

CYUL        CYLU          CUYL      CULY       CLUY     CLYU

YCUL            YUCL             YULC        YLCU       YLUC     YCUL

  • There are 4 different letters and there are 24different arrangements.
  •  In the name “SAM” (Investigating the different number of arrangement in a word with 3 different letters)

           SAM        SMA       MSA      MAS      ASM     AMS

  • There are 3 different letters in this name and 6 different arrangements.
  • Investigating the number of arrangements in a two-letter word with 2 different letters.

  JO         OJ

There are 2 different letters in this name and there are 2 different arrangements.

Table of Results

Number of Letters

Number of Different Arrangements

2

2

3

6

4

24

5

120

6

720

7

5040

  • From the table of results I have found out that a 2-letters word with individual letters has 2 arrangements, and a 3-letter word with individual letters has 6.
  • Taking for example a 3-letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4! Which is called 4 factorial that is the same as     4 x 3 x 2.

  • So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8-letter word.

 The formula for this is:      n! = a

  • Where n = the number of letters in the word and

a = the number of different arrangements.

Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, 3 letters repeated and 4 letters repeated.

  • This is a 3-letter word with 2 letters repeated
...read more.

Middle

PQMMM      PMQMM      PMMQM      PMMMQ

QPMMM      QMPMM      QMMPM      QMMMP

MPQMM      MPMQM        MPMMQ     MQPMM

MQMPM        MQMMP      MMPQM     MMQMP

MMMPQ       MMMQP      MMPQM      MMMQP

  • 5-letter word, 3 letters repeated, 20 different arrangements.
  • This a 4- letter word with 3 letters repeated

SMMM     MSMM     MMSM     MMMS

 4-letter word, 3 letters repeated, 4 different arrangements.

  • This is a 5-letter word with 4 letters repeated

  RRRRK      RRRKR      RRKRR     RKRRR     KRRRR

 5-letter word, 4 letters repeated, 5 different arrangements.

Table of Results

No of letters

(n)

No letters repeated

Same letter repeated 2x

(p)

Same letter repeated 3x

(p)

Same letter repeated 4x

(p)

Same letter repeated 5x

(p)

3

6

3

1

0

0

4

24

12

4

1

0

5

120

60

20

5

1

6

720

360

120

30

6

7

5040

2520

840

210

42

I have worked out that if I say 5! = 120, to find out how many different arrangements in a 3 letter word (3x2 x1=6) it would be 5! Divided by 6= 20, so, a 6-letter word with 4 letters (4x3x2x1=24) repeated would be 6! Divided by 24 = 30, as you can see in the “No letters repeated” column these are the numbers we are dividing by:

  • 2 letters the same =   n!

(2x1 = 2)    

  • 3 letters the same =   n!

( 3x2x1 = 6)

...read more.

Conclusion

       = 24      = 6 different arrangements                                                     (1x2) x (1x2)     (2a’s) x(2b’s)                4

A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y)

So: 1x2x3x4x5            (5 letter words) = 120  = 5 different arrangements

    (1x2x3x4) x (1)       (4a’s x 1b)             24

 A five letter words like abcde; this has 1 of each letter (no letters the same)

So: 1x2x3x4        = 24       = 24 different arrangements

 (1x1x1x1x1x1)       1

 A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's).

So: 1x2x3x4x5 =            120  =10 different arrangements

    (1x2x3) x (1x2)            12  

 This shows that my formula works:

n!   = The number of letters in the word

        x!y! = The number of repeated letters the same

Conclusion

I was able to find these formulas

  • The formula for this is:      n! = a

            Where n = the number of letters in the word and

a = the number of different arrangements

  • This formula is used to find out the different arrangements and possibilities from individual letters in a word.
  • 2 letters the same =   n!

(2x1 = 2)    

  • 3 letters the same =   n!

( 3x2x1 = 6)

  • 4 letters the same =   n!

(4x3x2x1 = 24)

  • 5 letters the same =   n!

(5x4x3x2x1 = 120)

  • 6 letters the same =   n!

                                 (6x5x4x3x2x1 = 720)

  • Finding similarities in letters the formula I found are shown above.

Deborah Taiwo

...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

    so on, until the number being times by is the number 1 itself. Then divide the number by the number of times the letter has been repeated. No. of = No. of letters X ( No. of letters - 1 )

  2. I am investigating the number of different arrangements of letters

    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

  1. Emma's Dilemma

    LUCY) and divide it by 6. 24 ? 6 = 4 different arrangements. Why you divide by 6: When finding the number of different arrangements for words that have one letter repeated 3 times, you take the number of different arrangements for a word which has the same amount of letters, but none repeated, and divide it by 6.

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

    Five letters: AABCD ACABD BAACD CAABD DAABC AABDC ACADB BAADC CAADB DAACB AACBD ACBAD BACAD CABAD DABAC AACDB ACBDA BACDA CABDA DABCA AADBC ACDAB BADAC CADAB DACAB Total: AADCB ACDBA BADCA CADBA DACBA 60 ABACD ADABC BCAAD CBAAD DBAAC ABADC ADACB BCADA CBADA DBACA ABCAD ADBAC BCDAA CBDAA DBCAA ABCDA

  1. Emma’s Dilemma.

    Four Letters: 'LUCY' I have already found the number of possible permutations for a name with four letters (Lucy) in part two. There are a total of 24 permutations, which shows the pattern widening out even more. Five Letters: 'KACIE' K = 1 A =2 C = 3 I =

  2. Investigating the arrangements of letters in words.

    is: n! Proof The formula is n! because it is a simpler way or writing N x N-1 x N-2...till N-x = 1. The formula is N x N-1 x N-2...till N-x = 1 because when a new letter is added the previous number (N-1 )

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