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• Level: GCSE
• Subject: Maths
• Word count: 1349

# Emma's Dilemma

Extracts from this document...

Introduction

Maths Investigation: Emma’s Dilemma

In my investigation I will investigate the number of different combinations a word can be put in. For example the word… Tim.

The letters in this word can be mixed up to show all the possible variations of combinations the letters can be put in. So a variation of the name Tim would be… Mit.

E.g.        TIM,   ITM,   MIT,

TMI,   ITM,   MTI.            …this shows all the possible combinations the letters can be put

into.  A total of 6 different combinations can be achieved.

I will begin by investigating the name LUCY.  I will work out all the possible letter combinations that can be produced from this name.  I have chosen this name because it has no letters the same and I first intend to investigate words with no letters repeated before perhaps moving on to that situation.

Total = 24 variations.

I will try the same with a 3-letter name to see if there is some sort of pattern.

Total = 6 variations

I can not yet see any sort of connection yet other than they are both even numbers, I will do the same thing with a 2-letter name.

Total = 2 different arrangements.

Middle

The table below shows the results I have collected and the number of different arrangements for a 5,6 and 7 letter word that I have predicted.

To write my equation so as to work out an x length word with no repeated letters I have formed an equation using letters to symbolise the various digits. A short version of the method I have used can be shown by this equation:                                                                  n! = a

Where n = the number of letters in the word.

And     a = the total number of re-arrangements possible.

! is a mathematical term meaning factorial. So 5! Is the same as 4*3*2, and 4! Is the same as 4*3*2. This is the method I have used, so to save time I have used the ! in stead of writing out the whole sum.

Some words have 2 letters or more the same such as...EMMA.  I will now compare how having a repeated letter effects the total quantity of arrangements possible with these  words, which have a repeated letter in. I would like to find out whether the same equation is correct for this or whether it must be modified in some way.

A 4 letter word with 1 repeat in…

A total of 12 different arrangements.

Having a repeated letter obviously does effect the total amount of different arrangements possible as in a 4-letter word with no repeats in there were 24 different arrangements whereas as with this repeated letter there are only half the amount,12. A word which contains 2 of the same letters has half the different arrangements of that with no repeats. In accordance to this a 3 length word with 1 repeated letter should have 3 different arrangements because this is half of the total of the same length word with no repeats.

NOO

ONO

OON

Total =The 3 letter word also follows this predicted trend.

I can now produce a table to show the predicted total amount of arrangements for any length word with 1 repeat in.

## Table of Results

 No of letters(n) No letters repeated Same letter repeated 2x(p) 3 6 3 4 24 12 5 120 60 6 720 360 7 5040 2520

Conclusion

 No of letters(n) No letters repeated Same letter repeated 2x(p) Same letter repeated 3x(p) Same letter repeated 4x(p) Same letter repeated 5x(p) 3 6 3 1 0 0 4 24 12 4 1 0 5 120 60 20 5 1 6 720 360 120 30 6 7 5040 2520 840 210 42

The equation:  n!

p!

…only works if there is just one letter repeated.  So my equation would work to find out the total number of different arrangements for a combination like…mmab but not for mmbb…as this has 2 pairs of the same letters in. I can modify my equation slightly to account for this however…

n!

p!*m

Using this equation a 4-letter word with 2 letters repeated once should have 6 different arrangements…

My equation seems to work I have managed to predict the amount

arrangements , but to make sure I am not wrong I will try a 4-                                       letter word                               letter word  with 3 of the same letters in…

I predict this word will have a possible…

4!

3!*1 = 4 different arrangements

Total arrangements = 4

This is also more proof that my theory works so I can find the number of arrangements of any word with any number of repeats in using my final equation.  I find this to be satisfactory evidence to more a less prove my theory correct.  Therefore I have solved Emma’s dilemma and found out the formula.

Erin Baker        Page         5/9/2007

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