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Emma's Dilemma

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Introduction

AMDG        Maths Coursework        18/06/01

Emma’s Dilemma

Arrangements for Emma:

emma

emam

eamm

mmae

mmea

meam

mema

mame

maem

amme

amem

atmm

There are 12 different arrangements for Emma’s name.

Arrangements for Lucy:

lucy

ucyl

cylu

ycul

luyc

ucly

cyul

yclu

lcuy

ulcy

culy

yulc

lcyu

ulyc

cuyl

yucl

lyuc

uycl

clyu

ylcu

lycu

uylc

cluy

yluc

There are 24 different arrangements for Lucy’s name: Double the amount of Emma’s name.  I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For example there are 6 arrangements with Lucy beginning with L, and 6 beginning with u and so on. 6 X 4 (the amount of letters) gives 24.

My strategy for working out the arrangements was using systematic patterns. For Emma’s name, I started off with the letter E and once I had found all the different arrangements beginning wit E, I moved on to M and then A.  I used the same principle for Lucy’s name.

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Middle

Akmr

Amkr

Amrk

Rmak

Rmka

Rkma

Rkam

Ramk

Rakm

Kmar

Kmra

Krma

Kram

Karm

kamr

There are 24 different arrangements for Mark’s name.

As I have found that there were 24 arrangements for a 4 letter word with all different letters and that there were 6 ways beginning with one of the letter, I predict that there will be 120 arrangements for JAMES, 24 for j, 24 for a, 24 for m; etc. 120 divided by 5 (number of letters) equals 24.  Previously in the 4-letter word, 24 divided by 4 equals 6, the number of possibilities there were for each letter.

My prediction was correct as I have calculated the answer and it matches my prediction.

For Aaron, where

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Conclusion

This is 4 letters with two the same

xxyy

Xyxy

Yxxy

Xyyx

Yyxx

yxyx

There are 6 arrangements

What if I had xxxyy?

x=a, y=b

aaabb

aabab

aabba

ababa

abaab

abbaa

bbaaa

baaab

babaa

baaba

There are 10 different arrangements for this instance.

What if I had an arrangement of xxxxy?

x=a, y=b

aaaab

aaaba

aabaa

abaaa

baaaa

There are 5 different arrangements for this instance.

If I go back to xxxyy; there are 3 x’s and 2 y’s in a total of 5. As each letter has its own number of arrangements i.e. there were 5 beginning with x, and 5 beginning with y, I think that factorial has to be used again. Also in a 5-letter word there are 120 arrangements and 24 arrangements (120 divided by 5) for each letter. I came up with; the number of total letters factorial, divided by the number of x’s, y’s etc factorised and multiplied.

Formula equals n=!

...read more.

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