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Emma's Dilemma

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        Different combinations of letters for the name Emma:

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

Different combinations of letters for the name Lucy:

  1. LUCY                              
  1. ULCY
  1. CLYU
  1. YCLU          
  1. LUYC            
  1. ULYC          
  1. CLUY                    
  1. YCUL
  1. LCUY
  1. UCYL
  1. CULY
  1. YULC
  1. LCYU
  1. UCLY
  1. CUYL
  1. YUCL
  1. LYCU
  1. UYCL
  1. CYLU
  1. YLCU
  1. LYUC
  1. UYLC
  1. CYUL
  1. YLUC

The method that I have used to find the number of combinations of letters for these two names could be called manual. I have taken one letter from the name, and added to it the different combinations of the remaining three letters. I did that for every letter in the word, and that is how I reached 24 combinations. I have noticed that both names have 4 letters in them, but the name “Emma” only have 12 combinations. The reason for this is that the word has two letters that are the same (MM). Later on in this coursework I will state, and explain the formulae that can be used to find the number of combinations of letters for words that have two or more of the same letters in them.  

I have also noticed that there are only 6 possible combinations per letter.

...read more.


  1. NHOJ

I have spotted a pattern with the different combinations.


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

  1. AMMME
  1. MMAME
  1. MEAMM
  1. EMMMA
  1. AMMEM
  1. MMEMA
  1. MAEMM
  1. EAMMM
  1. AMEMM
  1. MMMAE
  1. MEMMA
  1. EMMAM
  1. MAMEM
  1. MMMEA
  1. MAMME
  1. EMAMM
  1. MEMAM
  1. MMEAM
  1. MMAEM
  1. AEMMM
...read more.


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!


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}.

...read more.

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