# Emma's Dilemma.

Extracts from this document...

Introduction

Emma’s Dilemma

This investigation is set in order to find a formulae to solve how many arrangements you can get for any word, no matter how many letters or how many letters are repeated. First of all a formulae will be found for finding out how many arrangements can be found for the name LUCY. When this is found, I will then find out how many arrangements one can get from the name EMMA. After finding how many arrangements there are, I will put them into a formulae and see if there is any relation between the first part and the second part of this investigation. Once I have tried to make a link between these sections, I will move onto part 3, and decipher a way that you can work out how many arrangements there are for any word.

Part 1 – Lucy

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, UCLY, UYLC, ULYC, UCYL, UYCL,

YCUL, YUCL, YULC, YLCU, YLUC, YCLU, CYLU, CYUL, CUYL, CULY, CLUY, CLYU.

Middle

5

120

6

720

7

5040

Using this equation I can work out that for a 10 letter word there would be 3628800 arrangements for the word.

Part 2

Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, this time using EMMA.

EMMA, AMME, AMEM, EMAM, AEMM, EAMM, MMEA, MMAE, MEMA, MAME, MEAM, MAEM

There are 12 different ways of arranging EMMA. This is a four letter word with two letters repeated. With this in mind, I notice that on the first investigation, I had a four letter word with no letter repeated, now I have a four letter word with two letter repeated. In the first experiment I had 24 different ways and now in this one I have 12. Therefore I notice that 12 is half of 24, which would make sense as two of the letters are repeated in which would half the amount of ways you can rearrange the word. Therefore using part of section 1, I can work out a new equation.

n! / a! = x (number of letters factorial divided by the number of letters repeated equals the number or arrangements.)

Conclusion

1

5

120

60

20

5

6

720

360

120

30

Part 3

Now I have two equations worked out I have to work out the equation in which will work out the amount of arrangements for any word.

Using a four letter word with two letters repeated this is what I get:

AABB, BABA, BBAA, ABAB, ABBA, BAAB

This is 6 different arrangements. Now all I have to do is link it to the other two parts of this investigation. Doing the first part of the word would be like this, 4! / 2! = 12, then there are two letters left in the word, and that would decrease the amount of ways there are by two, there that would make my equation, 4! / (2!) (2!) = 6.

To test my new equation I have picked a 5 letter word with 2 letters repeated and 3 letters repeated.

FFFLL, FFLFL, FFLLF, FLFLF, FLFFL, FLLFF, LLFFF, LFFFL, LFLFF, LFFLF

There are 10 arrangements of these letters, so using my equation....

5! / (3!) (2!) = 10

Therefore my final equation is

n! / (a!)(b!)(c!)……(z!)

This means number of letters in the word divided by the number of letters repeated.

E.g. A five letter word like tttts; this has 4 t's and 1 s

So: 5! / (1x2x3x4) x (1) = 24 arrangements

A 7 letter word; aabbcdd

7! / (1x2) (1x2) (1x2) (1) = 630 arrangements.

So in conclusion, my final formulae is

n! / (a!)(b!)(c!)…… (z!)

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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