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# Emma's dilemma The different ways of arranging letters for Emma's name

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Introduction

Emma’s dilemma

There are different ways of arranging letters for Emma’s name the combinations for Emma’s name are

EMMA        MMAE        MMAE        EAMM        AEMM        MEMA

MAME        MEAM        MAEM        AMME        AMEM        EMAM

There are 12 combinations for a 4 letter word with 2 repeats Emma has a friend called Lucy. Lucy wants to find how many different combinations her name makes

There are 24 different combinations for a 4 letter word with no repeats

1 Letter

A

2 Letters

AB BA

3 Letters

ABC ACB BAC BCA CAB CBA

4 Letters

 No. letters combinations 1 1  = 1 x 1 2 2 = 1 x 2 3 6 = 2 x 3 4 24 = 6 x 4 5 ? = 5 x 24

The rule is times

Middle

EABCD

EACDB

EACDB

EACBD

EDCAB

EDCBA

EDBAC

EDBCA

EDACB

EDABC

ECBAC

ECBDA

ECDBA

ECDAB

ECABD

EBACD

EBCDA

EBDCA

EBDAC.

I then swapped around D with E for example

EABCD DABCE

EACDB DACEB

I done this until I had 124 different combinations or 24 different combinations for each starting letter this method went as good as the over one because it was confusing and took longer.

2 repeated letters

1 letter

No combinations

2 letters

AA

3 letters

AAB ABA BAA

4 letters

AABC AACB ABCA ABAC ACAB ACBA BAAC BCAA CBAA CABA CAAB

 No. letters combinations 1 No combinations 2 1 = 2 x 1 divided by 2 3 3 = 3 x 2 x 1 divided by 2 4 12 = 4 x 3 x 2 x 1 divided by 2 5 ? = 5 x 4 x 3 x 2 x 1 divided by 2

Conclusion

3

No combinations

4

4 = 4 x 3 x 2 x 1 divided by 24

5

? = 5 x 4 x 3 x 2 x 1 divided by 24

Using N factorial and dividing by six I can predict that 5 will have 5 different combinations because the rule is N! Divided by 24.

CONCLUSION

Therefore, the rule for finding out number combinations is the number of letters factorial. So 4 would be 4! (4x3x2x1) which = 24 number of combinations. The rule for finding the combinations with a double or treble ECT. Is the number of repeats factorial divide the number or letters factorial so a 4 letter combination with two repeated letters for example abbc would be 4! (4x3x2x1) which equals 24 divided by 2! (2x1) which equals 2 so that means there will be 12 combinations for a 4 letter combinations with 2 repeats.  Anther example could be an 8 letter combination with 4 repeated letters for example abcdeeee would be 8! (8x7x6x5x4x3x2x1) which equals 40320 dived by 4! (4x3x2x1) which equals 24. 40320 divided by 24 = 1680 so for an eight letter combinations with 4 repeats there are 1680 combinations

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