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• Level: GCSE
• Subject: Maths
• Word count: 1055

# Emma's Dilemma

Extracts from this document...

Introduction

Emma’s Dilemma

The Problem

Emma is looking at the different arrangements of her name.

1)        Find the different arrangements of Emma’s name.

2)        Emma has a friend named Lucy; find the different arrangements of her         name.

3)        Investigate the number of different arrangements of the letters of names         you have chosen.

Investigating the Problem

To investigate this problem I am going to look at the different arrangements of a four letter name with two letters the same and a four letter name with all different letters. I am then going to try three and five letter names, with two letters the same and all different. I am then going to look at the results and predict the next set. I will then move on to three and four letters the same.

Four Letter Words

One/Two Letter Words

We know that a one letter word will only have 1 combination. For two letters we know that it only has 2 different combinations and 1 for both letters repeating:

1. A
1. AB
2. BA
1. AA

Three Letter Words

Five Letter Words

James is a five letter word with different letters.

Middle

5

120

60

From the results we can already see a pattern. The combination for all different letters divided by two equals that of two letters the same in a word.

Prediction

By looking at the results I have found a pattern. The table below highlights the pattern.

 1 1 - 2 2 1 3 6 3 4 24 12 5 120 60

To predict the result for 6 letters:

120 X 6 = 720

To check this:

Let’s take ABCDEF. We know that by putting the A at the front of the word, we are only changing a five letter word. The results for a five letter word with different letters in 120.

120 X 6 = 720

This is the same sum as the pattern that I found. So, we can predict the results for the rest of the numbers by using this rule.

Factorial

The factorial button on a calculator [ ! ] proved very helpful in finding out the rule to Emma’s Dilemma.

Factorial means multiplying itself by all the numbers lower than itself until reaching 1.

e.g.        3!        =        3 X 2 X 1 = 6

4!        =        4 X 3 X 2 X 1 = 24

By using the factorial rule, the number of combinations is easier to work out.

Conclusion

( n ! / 2 ) / 3                        simplified        n!/6         because 2 X 3 = 6

I predict that for four letters the same the formula would be:

( ( n ! / 2) / 3 ) / 4                simplified        n!/24        because 2 X 3 X 4 = 24

The Formula

There is in fact a link between the formulas shown above where you divide n! by 2, 6, 24 e.t.c. and the factorial numbers itself. By looking at the numbers we can see that 2, 6, 24 is in fact 2!, 3!, 4! and so on.

( n ! / 2 ) / 3                 =                n!/6                =        n!/2!

( ( n ! / 2) / 3 ) / 4        =                n!/24                =        n!/3!

Let’s put this in a table:

 n n!/1! n!/2! n!/3! Letters All different 2 the same 3 the same

Notice that the 2! or 3! that n! is divided by equates to the number of letters  the same in each word. By using the information on this page we can find a general formula for the ‘dilemma’.

N!

L!

Where…

N        is the number of letters in the word;

L        is the number of letters the same.

This formula is a general rule for all combinations containing any number of letters with a number of letters the same (excluding combination of letters that have more than one letter repeated, i.e. AABB cannot use this formula).

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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