# I am investigating the number of different arrangements of letters in a word.

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Introduction

David Beeney

I am investigating the number of different arrangements of letters in a word.

The name Emmahas 12 different combinations.

24 combinations

- LUYC
- LCYU
- LCUY
- LYCU
- LYUC
- UCLY
- UCYL
- ULCY
- ULYC
- UYLC
- UYCL
- CUYL
- CULY
- CLUY
- CLYU
- CYUL
- CYLU
- YCLU
- YCUL
- YULC
- YUCL
- YLCU
- YLUC

The name Lucyhas 4 letters with 24 combinations.

2 letters

- Jo 2 combinations
- Oj

The name Jo has 2 letters with 2 combinations

3 letters

- Sam 6 combinations
- Sma
- Ams
- Asm
- Mas
- Msa

The Name Sam has 3 letters with 6 combinations

5 letters

- D(iego) =24 +
- I (dego) =24 +
- E (digo) =24 +
- G (dieo) =24 +
- O (dieg) =24 +

=120

Total no. Of letters | No repeated letters |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

In a 4-letter name such as Lucy, the way to work out the number of combinations is to do 4 x 3 x 2 x 1 or 4 factorial (4!). . Factorial is a number multiplied by the previous consecutive numbers.

Middle

With this formula in mind I can predict the number of arrangements for an 8-letter name. 8! (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) =40320

When there are repeated letters like in EMMA there are 12 combinations.

EMMA

AMME

AMEM

EMAM

AEMM

EAMM

MMEA

MMAE

MEMA

MAME

MEAM

MAEM

Emma has 2 M’s in it which means that the formula that I used for LUCY would not work. Emma is a 4-letter word with 2 letters repeated with 12 different combinations. I investigated other word combinations with 1 or more letters the same.

DAD

DDA

ADD

Dad is a 3-letter word with 2 letters repeated and 3 different combinations.

IDMMM

IMDMM

IMMDM

IMMMD

DIMMM

DMIMM

DMMIM

DMMMI

MIDMM

MIMDM

## MIMMD

MDIMM

MDMIM

MDMMI

MMIDM

MMDMI

MMMID

MMMDI

MMIDM

MMMDI

MMMDI is a 5-letter word with 3 letters repeated and has 20 different arrangements.

SIII

ISII

IISI

IIIS

IIIS is a 4-letter word with 3 letters repeated and 4 different arrangements.

BBBBD

BBBDB

BBDBB

BDBBB

DBBBB

DBBBB is a 5-letter word with 4 letters repeated and 5 different arrangements.

Conclusion

Number of combinations

4

24

2x’s, 2y’s

6

5

120

3x’s, 2y’s

10

From this table I concluded that in order to predict the amount of combinations with 2 or more letters the same you do for example, for a 4 letter word with 2 x’s and 2 y’s you do number of letters factorial divided by the number of x’s factorial multiplied by the number of y’s factorial. So if I substitute figures into the figures form my table I get 4! (24)/2!(2) x 2!(2) =6

I am now going to try and predict the amount of combinations for Mississippi using the formula I have worked out above. It is an eleven letter word with 1 m 4 s’es, 4 i’s and 2 p’s. So 11! =39916800. I need to divide this by 1! =1 multiplied by 4! =24 multiplied by 4! =24 multiplied by 2! =2 which all equals 1152. So my final sum is 39916800/1152=34650.

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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