# Emma&#146;s Dilemma

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Introduction

Emma’s Dilemma

In my investigation I am going to investigate the number of different arrangements of letters in a word.

e.g.

Tim

Is one arrangement.

Mit

Is another.

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

LUCY UCYL CYLU YULC

LUYC UYLC CYUL YUCL

LCYU UCLY CLUY YLUC

LCUY UYCL CLYU YLCU

LYCL ULCY CUYL YCLU

LYLC ULYC CULY YCUL

There are 4 different letters and there are 24 different arrangements.

As you can see, the method i have used to make sure that I have found the correct amount of arrangements is that there are 4 letters and each letter should have 6 different arrangements. Two with the same second letter, then another two with a different second letter, and then the last two also have the same second letter.

SAM

SMA

MSA

MAS

ASM

AMS

There are 3 different letters in this name and 6 different arrangements. Again, as you can see there a pattern to how I have worked out the correct amount of arrangements, there are 2 different letters after the first letter, which means there are 2 different combinations for each first letter.

Middle

3

6

4

24

5

120

6

720

7

5040

From the table of results you can see that a 2 letter word has 2 arrangements, and a 3 letter word 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 which 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.

EMMA

AMME

AMEM

EMAM

AEMM

EAMM

MMEA

MMAE

MEMA

MAME

MEAM

MAEM

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

MUM

MMU

UMM

3-letter word, 2 letters repeated, 3 different arrangements.

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.

SMMM

MSMM

MMSM

MMMS

Conclusion

In the above example there are 3 x's and 2 y's

As each letter has its own number of arrangements i.e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again.

As before, the original formula:

n! = the number of letters in the word

p! = the number of letters the same

From this I have come up with a new formula The number of total letters factorial, divided by the number of x's, y's etc factorised and multiplied.

For the above example:

A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)

So : 1x2x3x4

(1x2) x (1x2)

= 24

4 = 6 different arrangements

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

So: 1x2x3x4x5

(1x2x3x4) x (1)

= 120

24 = 5 different arrangements

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

So : 1x2x3x4

(1x1x1x1x1x1)

= 24

1 = 24 different arrangements

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

So : 1x2x3x4x5

(1x2x3) x (1x2)

= 120

12 = 10 different arrangements

This shows that my formula works:

n! = the number of letters in the word

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

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

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