# Emma's dilemma.

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Introduction

Maths CourseworkCharlotte Nellist

We were given the name ‘Emma’ and asked to look at it and see how many different ways the letters could be arranged in. But ‘Emma’ has a repeat in it; it has two m’s. I decided to start off with the simpler option by not having any repeats in the words, also I thought I would go through first doing a 2 letter word then 3 and then 4… and then I would put all the information into a table then try and work out repeats.

JO - jo

- oj

SAM - sam sam

- asm I put them into a bit more sma

- mas order so it was easier to understand. ams

- msa I did this by doing the ‘s’ first asm

- ams then the ‘a’ and then the ‘m’. msa

- sma mas

TONI – toni - nito

- toin - niot

- tion - ntio ← There are 6 different arrangements for each letter.

- tino - ntoi So altogether for a 4 letter word there are 24

- tnoi - noti different ways.

- tnio - noit

------- -------

- onti - itno

- onit - iton

- otni - iotn

- otin - iont

- oitn - into

- oint - inot

KIREN - kiren - keinr - irenk - ierkn - ikern

- kirne - keirn - irekn --------- - iknre

Middle

--------- --------- --------- --------- ---------

- senail - slaein - sielna - snliae - sainel

- senali - slaeni - sielan - snliea - sainle

- senlai - slaien - sienal - snlaie - saieln

- senlia - slaine - sienla - snlaei - saienl

- senile - slanie - siealn - snleai - sailne

- senial - slanei - sieanl - snleia - sailen

--------- --------- --------- --------- ---------

- sealin - sleina - silnae - sniael - saneil

- sealni - sleian - silnea - sniale - saneli

- seailn - slenai - silane - snieal - sanlie

- seainl - slenia - silaen - sniela - sanlei

- seanli - sleani - silean - snilea - sanile

- seanil - sleain - silena - snilae - saniel

There are 120 different ways of arranging the word ‘selina’ beginning with the letter ‘s’. There are 6 letters in the word, so 6 x 120 = 720. Which is how many ways I predicted it would have.

A formula can be made to predict the number of different combinations for any number of letters in a word. If the number of letters is n then the formula will be n(n-1)(n-2)…

This works out because there are n possible

Conclusion

- adddcb - cadddb - dadcbd - dcdbad - dddacb

- bacddd - cbaddd - daddcb - dcddba - dddbac

- badcdd - cbdadd - daddbc - dcddab - dddbca

- baddcd - cbddad - dadbdc - dcdadb - dddcab

- badddc - cbddda - dadcdb - dcdbda - dddcba

A 6-letter word with one letter repeated 3 times has 120 different arrangements. So the formula is correct.

Then I checked that it would work if in a word there were 2 letters repeated twice.

AABB - aabb

- abab

- abba

- baab

- baba

- bbaa

ABBCC - abbcc - bbacc - bacbc - bccab - cbbca

- abcbc - bbcac - baccb - bccba - cbcab

- abccb - bbcca - bcabc - cacbb - cbcba

- accbb - cabbc - bcacb - cbabc - ccabb

- acbcb - cabcb - bcbac - cbacb - ccbab

- acbbc - babcc - bcbca - cbbac - ccbba

Number of letters | Number of arrangements |

4 | 6 |

5 | 30 |

4! = 24

24

6 = 4

The factors of 4 are 2 and 2

Which are both letters repeated twice,

n(n-1)(n-2)(n-3)…

(x1(x-1)(x-2)…*x2(x-1)(x-2)…)

So for a 5-letter word with 2 letters repeated twice the equation will be,

5(5-1)(5-2)(5-3)(5-4)

(2(2-1)*2(2-1))

Or on a calculator,

5!

(2!*2!) = 30 different arrangements

With all this information the total number of combinations can now be calculated for any word with any number of letters with any number of repeated letters too. All these formulae have a name, factorial and a notation, n!.

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