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• Level: GCSE
• Subject: Maths
• Word count: 1567

# Emma's Dilemma.

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Introduction

Emma’s Dilemma

I am trying to find out the different arrangements of letters in a word with so many letters.

Also I am going to try and find out a way of finding these mathematically and also to try and find a formula which will work with my theory.

My way of working these out is by a simple method.

First I used the name Lucy, this is because all the letters in this name are different. So it will be easier. Then we used this method of:

1. I would keep the first letter the same all the time.
1. Then I would keep the second letter the same.
1. Then I would rotate the other letters around to make different arrangements.
1. After I have done this I would change the second letter to a different in the word but not the first letter.
1. Then I would do the same as in step 3 where I would rotate the last 2 letters, such as below.
1. Do the same but with the same first letter but a different second letter that you haven’t already used.
1. Once I have done this I have found all the different arrangements with the first letter.

Middle

To get the amount of arrangements for a 4 lettered word it would be 6x4, but to get the 6, you need the calculation for the arrangement before which was 1x2x3.

So I then substituted the answer for the calculation which was then:

6 x 4 = 24                        6 = answer

Substituted                        1 x 2 x 3 = Calculation

1 x 2 x 3 x 4 = 24

So it is still the same calculation as before except I have changed the answer with the calculation.

This works the same for the next one,

For a word with 5 letters, to get the amount of arrangements before it would be:

24 x 5

But to get the 24 you would need to do 6 x 4 or 1 x 2 x 3 x 4.

So for this calculation it would be:

1 x 2 x 3 x 4 x 5 = 120

Now I am going to try to find all the different arrangements of the letters in a word with 2 of the letters being the same.

 EMMAEMAMEAMM MEAMMAEMMEMAMMEAMMAEMAME AEMMAMMEAMEM

There are 4 letters, 2 being the same and there are 12 different arrangements.

## MUM

MMU

UMM

There are 3 letters, 2 being the same and there are only 3 different arrangements.

BB

There are only 2 letters the same and there are 1 arrangements.

 No. Of letters No letters repeated Same letter repeated 2x 1 1 Not possible 2 2 1 3 6 3 4 24 12

Conclusion

 No. Of letters No letters repeated Same letter repeated 2x Same letter repeated 3x 1 1 Not possible Not possible 2 2 1 Not possible 3 6 3 1 4 24 12 4 5 120 60 20

But if I make all the C’s different I get 24 arrangements. But if I pair up all the arrangements that are the same, as before, I only get 4 arrangements.

1 CCCA

2 CCAC

1 CCCA

2 CCAC

3 CACC

3 CACC

1 CCCA

2 CCAC

1 CCCA

2 CCAC

3 CACC

3 CACC

2 CCAC

1 CCCA

2 CCAC

1 CCCA

3 CACC

3 CACC

4 ACCC

4 ACCC

4 ACCC

4 ACCC

4 ACCC

## 4 ACCC

So if I put all the 1’s together, all the 2’s together, all the 3’s together and all the 4’s together, I get 4 arrangements.

CCCA
CCAC
CACC
ACCC

Now I am left with the 4 of the original 24 arrangements so the formula for this would be,

.n! / 6

It is divided by 6 because there was six of the same arrangements. For example the ACCC was repeated 6 times by,

ACCC

ACCC

ACCC                        n! / 6

ACCC

ACCC

ACCC

It is also six becausein a 3 lettered word you can only get 6 different arrangements, and towork out the six you do 3! So the formula can now be

n! / 3! = a

So now I have found an overall formula:

n! /x! = a

n = the number of letters

x = the number of letters repeated for n

a = the different number of arrangements.

This works because for a 4 lettered word with 3 letters the same repeated it would be:

4! / 3! = 4

And it also works for an 8 lettered word with 2 letters repeated:

8! / 2! = 20160

This shows that my theory works.

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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