Emma's Dilemma

Therefore the formula for any group of different letters is:

n!

n = the number of letters in the word

The definition of factorial is:

“The product of all the positive integers from 1 to a given number”

So to find out the number of combinations in a group of 4 different letters, I would have to:

1 x 2 x 3 x 4

The number ‘4’ is of the number of letters in the word.

However this formula does not work with groups of letters with some of the letters the same in it as a group with some letters the same in it would have a lower number of combinations. I will now find the number of combinations for groups of letters with group of the same letters in it. I will start off with a two letter word:

Two letters the same:

1. MM

So there is only one combination for a two letter word with two of the same letters.

Here is a table of a two letter word:

I tried my former formula out on these results but it did not work as I’d have to use the number of letters the same and the number of letter in the word both in the formula and in my formula I only have room for one number.

I will now find the number of combinations for a group of 3 letters with a group of the same letters in it. I will start off with a three letter word:

Two letters the same, ‘ITT’:

1. ITT
2. TIT
3. TTI

So there are only three different combinations for a three letter word with two of the same letters.

Three letters the same, ‘TTT’:

1. TTT

So there is only one different combination for a three letter word with three of the same letters.

Here is a table of a three letter word:

The formula does not work for this one either; therefore I will have to find a different to my previous one however it might have something to do with the previous formula. It might involve ‘factorial’ as I had to use it in the previous formula but to find this overall formula out I will have to work out the number of different combinations for groups of different numbers with a group of the same number of letters in it.

I will now find the number of combinations for a group 4 of letters with a group of the same letters in it. I will start off with a four letter word:

Two letters the same, ‘EMMA’:

1. EMMA
2. EMAM
3. EAMM
4. MAME
5. MAEM
6. MEAM
7. MEMA
8. MMEA
9. MMAE
10. AMEM
11. AMME
12. AEMM

So there are only twelve different combinations for a four letter word with two of the same letters.

Three letters the same, ‘EMMM’:

1. EMMM
2. MEMM
3. MMEM
4. MMME

So there are only four different combinations for a four letter word with three of the same letters.

Four letters the same, ‘MMMM’:

1. MMMM

So there is only one different combination for a four letter word with four of the same letters.

Here is a table of a four letter word:

The information above will be of better help once I have the results for the five letter words.

I will now find the number of combinations for a group 5 of letters with a group of the same letters in it. I will start off with a five letter word:

Three letters the same, ‘EMMMR’:

1. EMMRM
2. EMRMM
3. ERMMM
4. EMMMR
5. RMMME
6. RMMEM
7. RMEMM
8. REMMM
9. MMMER
10. MMMRE
11. MMRME
12. MRMME
13. MRMEM
14. MMERM
15. MMREM
16. MERMM
17. MREMM
18. MMEMR
19. MEMRM
20. MEMMR

So there are only twenty different combinations for a five letter word with three of the same letters.

Four letters the same, ‘MMMME’:

1. MMMME
2. MMMEM
3. MMEMM
4. MEMMM
5. EMMMM

So there are only five different combinations for a five letter word with four of the same letters.

Five letters the same, ‘MMMMM’

1. MMMMM

With these results I am able to create a table. Here it is:

With these results I am going to find out the formula for ‘Emma’s Dilemma’. The first thing I am going to do is for each one I know I have to put the number of letters in the group to factorial.

I will start off with an example. A four letter word with two letters the same. I know that the numbers of combinations in it are twelve so I have to work out how I got that result.

Four factorial:

4! = 24

I know I cannot multiply the number by anything because that would only make the number bigger. If I look at the result 12 and 4! = 24 I know that that is just 24 divided by 2.

Coincidentally, the number of letters in the same in the group is 2 so maybe the formula is the number of letters in the word divided by the number of letters the same.

First I have two try this formula out on my other results:

Trial formula:

n = the number of letters in the word

y = the number of the letters the same

n!

y!

Example:

Five letters in the word, three letters the same:

5!

3!

The answer to this is twenty. This is the same answer to the result I got. To insure my formula is completely full proof, I will have to try another example.

Example:

Three letters in the word, two letters the same:

3!

2!

The answer to this is three which is exactly what my results said. This proves my formula right.

But what about groups of letters with more than one group of the same letters within it?

For example a group like ‘AABBB’ would not receive an answer as it has more then one group of the same letters within it.

To find this out, again I have to find all the possible combinations for the group of letters.

Here are all the possible combinations for ‘AABBB’:

1. AABBB
2. ABABB
3. ABBAB
4. ABBBA
5. BABBA
6. BABAB
7. BAABB
8. BBABA
9. BBAAB
10. BBBAA

There are ten possible arrangements for a group of 5 letters, with a group of 3 letters the same and a group of 2 letters the same.

I will start off by using my former formula:

5!

3!

The answer to this is twenty. I have to find out someway to place the other number in the formula to get ten from twenty. Coincidentally, twenty divided by two, the other group of letters the same in the group of letters, is equal to the answer.

This would make my formula into this:

n = the number of letters in the word

y = the number of the letters the same

z = the other number of letters the same (if any)

x = the other number of letters the same (if any)        …etc

n!

y!z!x!

I believe this formula is correct but first I will have to try out again:

Example:

‘AABB’

1. AABB
2. ABAB
3. ABBA
4. BBAA
5. BABA
6. BAAB

The number of combinations for this example is 6. Now I have to try to see if it works with my formula:

4!

2! x 2!

The answer to this is the same as my result. This proves that my overall formula is correct and will work for any possible group of letters and will also get all of its possible combinations.