Emma's Dilemma

I have spotted a pattern with the different combinations.

LETTERS           POSSIBILITIES

1 letter                          1

              ×

                                                               1 × 2 = 2

                 

2 letters     =                 2      

                                             2 × 3 = 6

 

3 letters                        6

                                            6 × 4 = 24

                 

4 letters                        24    

                                            24 × 5 = 120

                     

5 letters                       120

I am definitely sure that his theory is correct, because I can check it by using a “factorial” function on my calculator. Factorial function is 1×2×3×4…..

It is marked with the following symbol “!”. Therefore 6! would equal to 1×2×3×4×5×6 which would equal to 720. Therefore, if we wanted to find a number of combinations for a word with 5 letters, we would have to enter 5! in our calculator, and we would get 120. This confirms the pattern from above.

N! = N × (N-1) × (N-2) × (N-3) × ... × 3 × 2 × 1   In this case N is a positive integer. Factorial is introduced as the number of ways that an original set can be ordered. For example, if an original set contains 3 items {A, B, C}, then the number of ways that you can order these three items (set elements) is 3! = 3×2×1 = 6. Those combinations are:  {A,B,C}{A,C,B}{B,A,C}{B,C,A}{C,A,B} {C,B,A}

Therefore, I have concluded that the basic formula for this could be:

Y = N!

N = number of letters in a word

Y = number of all possible combinations

I will now investigate the formulae that could work in case two letters are the same in the word. I will take my first name as an example for this task.

1. FILIP
2. FIPIL
3. FILPI
4. FIPLI
5. FIIPL
6. FIILP
7. FLIIP
8. FLPII
9. FLIPI
10. FPIIL
11. FPLII
12. FPILI

I will now investigate the word that has three letters in it that are the same.

I will investigate the following set of letters:   E M M M A

Now I will discover the formula for three letters that are the same in the word. I will tackle a 6-letter word.  AAABCD

1. BAAACD
2. BAAADC
3. BAADCA
4. BAACDA
5. BACDAA
6. BADCAA
7. BDCAAA
8. BDACAA
9. BDAACA
10. BDAAAC
11. BCDAAA
12. BCADAA
13. BCAADA
14. BCAAAD
15. BCDAAA
16. BDCAAA
17. BACADA
18. BADACA
19. BAACAD
20. BAADAC

So the formulae for three same letters in a word is the following

               N!

Y =  

               6

I will now try and make this formulae more advanced, and I will try to all the combinations:

The pattern is the following

N = number of letters in a word

X = number of possible combinations

From this table I have noticed that X equals the last X × N, so I would expect the formulae for 4 same letters in a word to be the following:

                N!

Y =

                24

I will try to confirm this formula in as simple way as possible, and I will choose a 5-letter word, in which 4 letters will be the same:

AAAAB

AAABA

AABAA

ABAAA

BAAAA

Y = n!/24 = (1×2×3×4×5)/24 = 5

This formulae does work

There is an extremely simple method of finding out the number of possibilities. It is very similar to finding the number of combinations when all the letters are different, but in this way, it is rehearsed. If you have a six-letter word with three letters that are the same, you need to divide the number with one, and than two, and then three factorial.

For example: If you have a set of letters such as:   AAABCD

First, you need to find the number of combination for the set if all the letters were different. Than you divide it with the number of letters that are the same, but, you divide it once for each letter. So first, you would divide the number with one factorial, than two factorial, and then three factorial.

6! = 720

720 ÷ 3! = 120

There are 120 possible combinations for the set of letters AAABCD.

The algebraic formulae that I would use for this is the following:

       N!

X!

N = the number of possible combinations if all letters were different

X! = the set of letters that are the same in the word

In the following situation, the formulae changes a bit, but it is based on the same principles:

AAABBCD

N!

P! × R!

In this case, N is again the number of combinations if all letters were different, but P, and R represent the different sets of letters.

A A A        B B    C D

                                       P = 3            R = 2

Therefore the solution in this case would be.

7! = 5040

P! = 3! × R! = 2!

5040

6 × 2

So, the possible number of combinations for the set of letters AAABBCD is 420.

I will test this formula on different sets of letters. I will deliberately choose a long number, in order to prove this theory.

AAAABBBCCD – a set of 10 letters

Step 1:   find 10!  =  3628800

Step 2:   divide it with A! (4!) = 151200

Step 3:   divide the previous result with B! (3!) = 25200

Step 4:   divide the previous result with C! (2!) = 12600

And the number of combinations you get will be 12600

Simplified method for this is:

                 10!                                       3628800

Y =                                       =                                                    =      12600

             A!×B!×C!                                      4!×3!×2!

For example, if asked a question such as how many different ways that one could select a subset of size 3 from an original set of size 5, the answer is (5!) divided by

(2! × 3!) = 10. Original set of size five:{A,B,C,D,E} and distinctly different subsets of size 3: {A,B,C}, {A,B,D}, {A,B,E}, {A,C,D}, {A,C,E}, {A,D,E}, {B,C,D}, {B,D,E}, {B,C,E}, and {C,D,E}.