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• Level: GCSE
• Subject: Maths
• Word count: 2211

# Emma&amp;#146;s dilemma.

Extracts from this document...

Introduction

## Edexcel Maths Course Work

CANDIDATE NAME: Akbar Miah

SCHOOL NAME     : London Islamic School/Madrassah

Emma’s Dilemma

In this piece of work I will be investigating all the different arrangement of words, and try to find formulas that will enable us to calculate how many different ways you can arrange various lettered words.

Example the word car is one way of arranging it and arc is another way of arranging it.

This coursework is divided into three sections

I will try to find the different arrangements of the letters of the following names:

Part 1: Lucy is a four-lettered word, all the letters are different.

lucy    ucly     cluy    yucl

lcyu    ucyl     cuyl    ycul

lycu     uylc    culu     yluc

lcuy     ulyc    cluy    yccl

lyuc    ulcy     culy    yclu

lyuc     uycl    cluy    ylcu

With Lucy’s name you can arrange it 24 different ways.

I could also calculate all the different arrangements of this word by only working out the number of times Lucy’s name can be arranged when it begins with the letter l, u, c or y, for example the word Lucy can be arranged 6 times when it begins with l, so because there are four letters you can multiply 6 by four, which will give you the number of times you can arrange the word Lucy, which is 24, and as we can see from above this theory is correct.

Part 2: Emma: Emma is also a four-lettered word but two of the letters are the same.

emma    mmea    mmae    amme

emam    maem    amme    amem

eamm    meam    mame    aemm

Middle

5 (1x2x3x4x5)

120

From this I found a formula to find how many ways a word could be arranged.

The formula I have found is A = n! (when there is no repeated letters), A is for arrangements, n! stands  for the factorial, which is n x (n-1) x (n-2) x (n-3)…until 1, so in 5 lettered word the arrangements will be 5 x 4 x 3 x 2 x 1 = 120, it is also the same thing if it is done backwards example of a five letter word, 1 x 2 x 3 x 4 x 5 = 120. For example if  I want to calculate how many ways a four lettered word can be arranged, this is how it will be done, 1 x 2 x 3 x 4 =24.

As we can see from above results are right, so now we are sure that this formula will work for any number lettered word.

I can see another formula that might work.

 Number of letters 2 3 4 5 Result of arrangements 2 6 24 120

From this I found out a formula,  the formula is N  X result for N- 1, here "n" stands for the number of letters in a  word , multiplied by the number of arrangements that can be made with a word that has a letter less. for example if I want to find the arrangements of a 5 lettered word, I will have to multiply 120 by 24, the reason why I will have multiply by 24 is because a four lettered word (no double leters) can be arranged only 24 different ways.

The formula is N x result for N-1. To see if it works, lets try with a 3 lettered word.

N x result for N-1

3   x                3-1 =2(because a 2 lettered word can be arranged twice)

3 x 2 = 6

Yes it works lets try with a 5-lettered word

N x result for N-1

5 x                  5-1= 24 (because a 5 lettered word can be arranged 24 different ways).

This proves this formula works.

But what about a 1 lettered word, the formula would not work. Example it is obvious a 1 lettered word can be arranged only one way, but the formula will give 0 as the answer.

N x (n – 1)

1  x ( 1 – 1)

1  x  0 = 0

One thing with the above formula is that if a word contains a double lettered word, the above formula cannot calculate the number of arrangements a word will have if it has more than two same letters.

Example the word abb, abb

bab

bba

It can be arranged three times, the above formula will be N x result for (n-1)

3 x                (3-1=2)

3x2=6

From this we can see that the above formula cannot calculate the number of arrangements for a word that has two or more repeated letters. So this means that the above formula will only work with a word that has no repeated letters.

Before finding a formula to calculate the number of arrangement for a word that contains double letters, I will have to find out the results for words containing double letters.

1. 2 letters of the same in a three-lettered word.

Ana

Aan

Aan

It can be arranged 3 times.

4 lettered word containing 2 letters of the same.

Emma  mmea  mmae  amme

Emam  maem  amme  amem

Eamm  meam  mame  aemm

As I mentioned before if a word has 2 identical letters it can be arranged less differently if the letters are different, example Lucy can be arranged 24 different ways and Emma can be arranged 12 different ways.

 NUMBER OF LETTERS ( 2 SAME ) NUMBER OF WAYS IT CAN BE ARRANGED 3 (1x2x3)=6divide by2 = 3 3 4 (1x2x3x4)= 24 divide by 2=12 12 5 (1x2x3x4x5) = 120 divide by 2 = 60 60

Conclusion

As above we know the word aabb  can be arranged 6 different ways, so now lets use the formula to see if the formula works. The formula for the above word would be

1 x 2 x 3  x 4 = 24 = 6

(1x2) x (1x2) =  4

The above formula does work as we can see from above. Now lets try with another word, which has five letters with containing four letters, the same with two pairs.

abbcc   bbcca    ccbba

abcbc    bbcac    ccbab

acbcb    bbacc    ccabb

accbb    babcc    cacbb

acbbc    bacbc    cabcb

abccb    baccb    cabbc

bccba   cbbca

bccab   cbbac

bcbac   cbabc

bcbca   cbcba

bcabc   cbacb

bcacb   cbcab

The formula would be for this, 1 x 2 x 3 x 4 = 120 = 30

1x (1x2)x(1x20)= 4

Yes, the formula does work.

If there is a word that has two letters the same and occurs more than two times, the formula is A = n!/ z! y! x!, and so on.

Now I cannot think of any problem that would arise in finding how many ways any words  can be arranged, also any word that wouldn’t fit in to any of the above formulas.

From this investigating I learnt many things, for example the more letters a word contains the more different ways it can be arranged, I learnt how to find formulas etc. I really enjoyed doing this investigation , and the main benefit I got from this investigation is that I will never have any problem in calculating how many different ways a word can be arranged.

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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