# Emma's Dilemma

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Introduction

Emma’s Dilemma

In this investigation, I will be attempting to find out the formula that would give me the number of possible arrangements for any group of letters, even if the group contains the same letters.

To find this out, I will be investigating the possible arrangements for different groups of letters and from there I will be then using the number of arrangements for each of these groups to find a formula.

I will start off with groups of different letters and find the formula for that.

The first group of letters I will start off with is the group ‘LUCY’:

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

Therefore for a group of four different letters there will be 24 different arrangements.

I will now attempt to find out the number of arrangements for a group of 3 different letters, ‘EMO’:

- EMO
- EOM
- MOE
- MEO
- OEM
- OME

So there are 6 possible different arrangements for any group of 3 different letters.

I am now going to try to find out the number of arrangements for a group of 2 different letters, ‘EM’:

- EM
- ME

Therefore, there are only two different arrangements for any two letter group of different letters.

I also

Middle

2

1

I tried my former formula out on these results but it did not work as I’d have to use the number of letters the same and the number of letter in the word both in the formula and in my formula I only have room for one number.

I will now find the number of combinations for a group of 3 letters with a group of the same letters in it. I will start off with a three letter word:

Two letters the same, ‘ITT’:

- ITT
- TIT
- TTI

So there are only three different combinations for a three letter word with two of the same letters.

Three letters the same, ‘TTT’:

- TTT

So there is only one different combination for a three letter word with three of the same letters.

Here is a table of a three letter word:

Number of letters the same | 1 | 2 | 3 |

Number of different combinations | 6 | 3 | 1 |

The formula does not work for this one either; therefore I will have to find a different to my previous one however it might have something to do with the previous formula. It might involve ‘factorial’ as I had to use it in the previous formula but to find this overall formula out I will have to work out the number of different combinations for groups of different numbers with a group of the same number of letters in it.

I will now find the number of combinations for a group 4 of letters with a group of the same letters in it. I will start off with a four letter word:

Two letters the same, ‘EMMA’:

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

So there are only twelve different combinations for a four letter word with two of the same letters.

Three letters the same, ‘EMMM’:

- EMMM
- MEMM
- MMEM
- MMME

So there are only four different combinations for a four letter word with three of the same letters.

Four letters the same, ‘MMMM’:

- MMMM

So there is only one different combination for a four letter word with four of the same letters.

Here is a table of a four letter word:

Number of letters the same | 1 | 2 | 3 | 4 |

Number of different combinations | 24 | 12 | 4 | 1 |

Conclusion

Here are all the possible combinations for ‘AABBB’:

- AABBB
- ABABB
- ABBAB
- ABBBA
- BABBA
- BABAB
- BAABB
- BBABA
- BBAAB
- BBBAA

There are ten possible arrangements for a group of 5 letters, with a group of 3 letters the same and a group of 2 letters the same.

I will start off by using my former formula:

5!

3!

The answer to this is twenty. I have to find out someway to place the other number in the formula to get ten from twenty. Coincidentally, twenty divided by two, the other group of letters the same in the group of letters, is equal to the answer.

This would make my formula into this:

n = the number of letters in the word

y = the number of the letters the same

z = the other number of letters the same (if any)

x = the other number of letters the same (if any) …etc

n!

y!z!x!

I believe this formula is correct but first I will have to try out again:

Example:

‘AABB’

- AABB
- ABAB
- ABBA
- BBAA
- BABA
- BAAB

The number of combinations for this example is 6. Now I have to try to see if it works with my formula:

4!

2! x 2!

The answer to this is the same as my result. This proves that my overall formula is correct and will work for any possible group of letters and will also get all of its possible combinations.

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