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# Emma's Dilemma

Extracts from this document...

Introduction

Emma’s Dilemma

Emma

Eamm

Emam

Emma

Maem

Meam

Mame

Mema

Mmae

Mmea

Aemm

Amem

Amme

Emma is a 4 letter name and 2 of the letters are the same.  Emma has a total of 12 possible combinations.

Lucy

Lcuy

Lcyu

Lucy

Luyc

Lycu

Lyuc

Ucly

Ucyl

Ulcy

Ulyc

Uycl

Uylc

Cluy

Clyu

Culy

Cuyl

Cylu

Cyul

Yclu

Ycul

Ylcu

Yluc

Yucl

Yulc

Lucy is also a 4 letter name, and all off the letters are different.  Because of this it has a total of 24 possible combinations, as opposed to the 12 which we found in Emma.

Ann

Ann

Nan

Nna

Ann is a 3 letter name and 2 of the letters are the same, giving us a total of 3 combinations.

Sam

Sam

Sma

Ams

Asm

Mas

Msa

Sam is a 3 letter word, and all the letters are different, giving us a total of 6 combinations.

JJ

JJ

Middle

 No. of letters All different 2 Same 3 Same 4 same 5 Same 2 letters 2 1 3 letters 6 3 1 4 letters 24 12 4 1 5 letters 120 60 20 5 1

The table above shows us, to get the total number of combinations when 2 letters are the same, is, divide the number of letters by 2 factorial.  To get the total number of combinations, when 3 letters are the same, is, divide the number of letters by 3 factorial.  To get the total number of combinations when 4 letters are the same, is, divide the number of letters be 1 factorial.  And to get the total number of combinations when 5 letters are the same, is, divide the number of letters by 5 factorial.

Conclusion

n!/x!y!

n is the number of letters !

x is the number of x’s !

y is the number of y’s !

For example:

5 letters – xxxxy = 5! / 4! x 1! = 120 / 24 = 5

4 letters – xxyy = 4! / 2! x 2! = 24 / 4 = 6

all different – vwxyz = 4! / 1!x1!x1!x1!x1!x1! = 24 / 1 = 24

All of these formula have been proved in previous examples, which shows that my formula does work.

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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