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# Emma's Dilemma

Extracts from this document...

Introduction

Michael Walker                                                                                                             Maths Coursework

Emma’s Dilemma

I have been asked to find out how many ways I can arrange the EMMA. I have then been asked to work out a formula for this without writing down every combination. I will now find out how many ways I can arrange the letters EMMA. One combination is

EMMA                 Another is

EMAM

EAMM

MMAE

MAME

MEMA

MAEM

MEAM

MMEA

AEMM

AMME

AMEM

They are 12 different combinations of EMMA. 4 letters but 3 different letters

I will now work out how many combinations they are with a 4 letters all different I will try PHIL one combination is

PHIL                        ILPH

PHLI                        ILHP

PIHL                        IHPL

PILH                        IHLP

PLHI                        IPLH

PLIH                        IPHL

HILP                        LIHP

HIPL                        LIPH

HPIL                        LHPI

HPLI                        LHIP

HLIP                        LPIH

HLPI                        LPHI

I have found out that they are 24 different combinations. 4 letters all of them different. I have worked that this is double to EMMA.

Middle

MPILH                        MIHLP

MPLHI                        MIPLH

MPLIH                        MIPHL

MHILP                        MLIHP

MHIPL                        MLIPH

MHPIL                        MLHPI

MHPLI                        MLHIP

MHLIP                        MLPIH

MHLPI                        MLPHI

My prediction was correct they are 24 combinations for every letter beginning they are 24 combinations so 24 times 5 (the number of letters) equals 120.

I have worked out a formula, with PHIL’s name I have found that the total number of combinations was 24 with 4 letters all different 1x2x3x4=24, and with MPHIL they are 5 letters all different, 1x2x3x4x5=120. I have also found an even quicker way to do this on a calculator, they is a factual button that looks like n! Or x!. All this simply does is saving you put 1x2x3x4x5 etc in you just put the number of letters and press the button.

Conclusion

MMMAA

MMAMA

MMAAM

MAMAM

MAMMA

MAAMM

AAMMM

AMMMA

AMAMM

AMMAM

I have found out that they are 10 different combinations.

I have worked out a logical formula, the number of letters factorial, divided by the number of A’s, M’s factorised. I will now test my formula;

For example I will work out MMMAA they are 5 letters 2 different I would work out, 5 factorised (the number of letters) equals 120 divided by 3 factorised (the number of M’s) multiplied by 2 factorised (the number of A’s) equals 12 so 120 divided by 12 equals 10.

A 4 letter word such as EMMA, this has 3 different letters

1x2x3x4 divided by 1x2 equals 24 divided by 2 equals 12

Below is a table showing the combinations for words with repeated letters;

 Number of Letters Number of M’s Number of A’s Number of combinations 3 2 1 3 4 2 2 6 5 2 3 10 6 2 4 15 3 3 0 1 5 4 1 5 5 1 0 24 7 4 3 168

I have tested and proven my formulas and they both 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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