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• Level: GCSE
• Subject: Maths
• Word count: 1840

# Emma's Dilemma.

Extracts from this document...

Introduction

G.C.S.E Maths Coursework

Emma’s Dilemma

For my G.C.S.E maths coursework I have been given the task to investigate how many times the letters of any name can be arranged. The method I am going to be using is to explore how many times the letters of short and long names can be arranged. Once I have got the hang of this I will investigate how many times short and long letter names with repeated letters can be arranged. Once I have completed this, I will create a general formula which could be used with any name with or without repeated letters.

 AJ JA

How many arrangements can be made from a name with two different letters like AJ?

AJ has a total of two arrangements.

How many times can a three letter name like JON be arranged?

Here are the arrangements:-

 JON NJO JNO NOJ OJN ONJ

From this table, we can see that there are a total of 6 arrangements.

 EMMA EAMM EMAM MEAM MEMA MAEM MMEA MAME MMAE AEMM AMEM AMME

Below is a table showing the arrangements for EMMA. This time there are 2 repeated letters. Let’s see how many times this name can be arranged:-

For EMMA, there are 12 possible arrangements. There are 2 different letters and two letters the same, and there is a total of 4 letters.

Middle

I will need a formula to help me find how many arrangements there are when a word has a lot of letters, repeated or not.

I am now going to find out how many arrangements there are for a five letter word with no repeated letters like SMART.

If I divide 24 by 6 it will give me an answer of 6, this is the number of arrangements a 3 letter word with no repeated letters has. If I multiply 24 by 5, it ought to give me the number of arrangements for SMART. Therefore I predict that there will be 120 different arrangements for SMART.

As there is no space for me to put a table with 120 arrangements in it, I will have to rely on my prediction until I make a formula and only then I can prove this prediction right or wrong. But below is a short table of how I would start of expressing how SMART should be arranged:-

 SMART SARMT SAMRT SRMAT SMRAT SARTM SAMTR

As the table gets larger, I can see that there is going to be 5 horizontal rows and 24 vertical columns. This will give me a grand total of 120 boxes which is the number of arrangements for SMART.

Conclusion

N!

A! X B! X C!

Whereas A, B and C= the different number of repeated numbers.

If I want to find out how many arrangements there are for XYYXXXYY by using my formula, it would be calculated by:

8!          =

4! X 4!

40320          = 70 arrangements

576

Therefore the arrangements for XXXXXYYXYYXY would be:-

12!          =

7! X 5!

479001600              = 792 arrangements

604800

Conclusion

As you can see from my method of making a formula, it did not turn out correctly. This is because my formula turned out to be exactly the same as the factorial formula. As there is no quadratic formula because it is a factorial, the only formula to work out how many arrangements there are in words with no repeated letters is N! This is because (N) (N+1) (N+2) is the same as 1 x 2 x 3 = 6.

Therefore the only formula to work out how many arrangements there are in words with no repeated letters is N!

My other formula which will work out how many arrangements there are in words with repeated letters is:-

N!

A! X B! X C!

To prove this formula, I am going to calculate how many arrangements the word Allan has:-

5!         =    30 arrangements

2! X 2!

As this answer is correct, I declare that my formula works

Therefore my two formulas are N! and

N!

A! X B! X C!

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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