Emma's Dilemma

I then noticed that the total number of combinations is the number of letters multiplied by the previous number of combinations.

Here is another example using the name TIM.

Total number of letters: 3

Previous number of combinations: 2

3x2=6

Here is another example using the name DAISY

Total number of letters: 5

Previous number of combinations: 24

24x5=120

After doing this I realised that I can work out the amount of combinations using factorial notation. This is when a number is multiplied by the previous consecutive numbers.

For example, using the letter 6 you would do this: 6x5x4x3x2x1

I realised this is because if I could find the total number of combinations by multiplying the total number of letters by the previous number of combinations it was the same as multiplying the total number letters by its previous consecutive numbers.

Concluding, the formula to find out how many combinations there are when all the letters are different is

Number of combinations = Total number of letters Factorial

Here are some results to prove this

Now I am going to investigate the number of combinations when 2 letters are the same. For this I will use the name EMMA

Here are all the combinations in the name EMMA

EMMA         EMAM          EAMM           MMAE

MMEA         MEAM          MEMA           MAME

MAEM         AMME          AMEM           AEMM

This shows that all 4 letter words with 2 letters the same will have 12 combinations. I noticed that this is half the amount of combinations there was when all the letters were different. I then investigated this to see if it was true with all words with 2 letters the same.

As you can see, all of the total number of combinations with 2 letters the same is half the amount of when all the letters are different.

Then I investigated the difference between the total number of combinations when two letters are the same with the total number of letters factorial.

From this you can see that the total number of combinations is half of total number of letters factorial

From that I came up with the formula

Number of combinations = Total number of letters Factorial/2

Now I am going to investigate names with a different number of repeated letters. Here are some examples.

AABB

All of the combinations

AABB                 ABAB          BAAB

AABA                 BABA          BBAA

There are 6 combinations in total, 3 beginning with A and 3 beginning with B

AAABB

All of the combinations

AAABB         AABAB         AABBA         ABABA          ABAAB

ABBAA         BBAAA         BAAAB         BABAA          BAABA

There are 10 combinations in total, 6 beginning with A and 4 beginning with B

AAAAB

All of the combinations

AAAAB         AAABA         AABAA

ABAAA         BAAAA

There are 5 combinations in total, 4 beginning with A and 1 with B

Because there are two sets of same letters the previous formula cannot be used. Instead, I have come up with another formula:

Total number of letters factorial/

Number of repeated letters factorial (1) x number of repeated letters factorial (2)

For example

A five letter word like AAABB

This has 3 As and 2 Bs

So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10

A four letter word like aabb

This has 2 As and 2 Bs

So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6

A five letter word like AAAAB

This has 4 As and 1 B

So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5

Five letter words like ABCDE

This has 1 of each letter

So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24