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• Level: GCSE
• Subject: Maths
• Word count: 1116

Emma&amp;#146;s Dilemma

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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

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)

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

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