Emma's dilemma.

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

All of the combinations with a * next to them are cancelled out because if they were not colour coded then they would look exactly the same as the combination above it. That is why there are twice as many LUCYs to EMMAs.

I am now going to investigate some other names that I have chosen.

2 letters

• Jo                        2 combinations
• Oj

3 letters

• Sam                        6 combinations
• Sma
• Ams
• Asm
• Mas
• Msa

5 letters

• R(obin)        =24        +
• O(rbin)        =24        +
• B(roin)        =24        +
• I(robn)        =24        +
• N(robi)        =24        +

=120

Here is a results table that I have put all of my results in.

After putting my results in a table I was able to predict the number of arrangements for 5 different letters by realising a pattern. I realised that with 5 different letters there are 24 possibilities beginning with each different letter which is shown in the table where you can see that to get 120 you had to times the number of arrangements for a 4 letter word with the number of letters for a 5 letter word.

So therefore I can also predict that there will be 40320 different arrangements for an 8 – letter word by working out 5040 x 8 = 40320.

I can see that the numbers of possibilities for different arrangements are going to increase as more different letters are used. So as a general formula for names with n number of letters all different I have come up with a formula, which is the number of letters multiplied by the previous consecutive numbers. For example with Lucy's name: 1x2x3x4 = 24. This is expressed as factorial. Factorial is a number multiplied by the previous consecutive numbers. Factorial notation is symbolised using an exclamation mark !

In a 4-letter word, to work out the amount of different arrangements you can do 1 x 2 x 3 x 4 = 24, which means that the formula for arrangements that have all different letters is n!

When there are repeated letters like in EMMA I could only find 12 combinations. This means that the formula that I used for LUCY would not work. I decided to make a new table so I could find out what would happen if you had letters repeated.

 

To make it a bit easier instead of using letters as such I will use x's and y's (any letter). I will start with xxyy:

Arrangements for xxyy:

x=a, y=b

This is a 4-letter word with 2 different. I have done this with;

aabb abab baab

aaba baba bbaa

There are 6 arrangements. What if I had xxxyy?

x=a, y=b

aaabb aabab aabba ababa abaab

abbaa bbaaa baaab babaa baaba

There are 10 different arrangements for this instance.

What if I had an arrangement of xxxxy?

x=a, y=b

aaaab aaaba aabaa

abaaa baaaa

There are 5 different arrangements for this instance.