# To investigate the number of different arrangements of letters in a different words

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Introduction

Aim: To investigate the number of different arrangements of letters in a different words

For Example:

To could be arranged as ot

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

LUCY

LUYC

LYCU

LYUC

LCYU

LCUY

ULCY

ULYC

UCLY

UCYL

UYLC

UYCL

CLYU

CLUY

CULY

CUYL

CYLU

CYUL

YLUC

YLCU

YULC

YUCL

YCLU

YCUL

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

SAM

SMA

MSA

MAS

ASM

AMS

There are 3 different letters in this name and 6 different arrangements.

TO

OT

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

Obviously there is only 1 arrangement for a word with one letter.

Number of Letters in a word | Number of Arrangements |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

From the table of results I have found out 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.

Middle

6

6x5x4x3x2x1

720

7

7x6x5x4x3x2x1

5040

8

8x7x6x5x4x3x2x1

40320

9

9x8x7x6x5x4x3x2x1

362880

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.

I am going to start with EMMA

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

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

RRRRK

RRRKR

RRKRR

RKRRR

KRRRR

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

When looking at the number of arrangements for Emma I realized that there were 12 and this was half of the number of arrangements for Lucy so I decided to look at the others. This also was the case for the 3 letter words such as Mum and Sam.

24/2 = 12 – The number of arrangements for Emma.

6/2 = 3 – The number of arrangements for Mum.

I started to look at what they had in common apart from the divide by two.

Conclusion

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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