Emma's Dilemma
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Introduction
Different combinations of letters for the name Emma:
 EMMA
 MEAM
 AEMM
 EAMM
 MMEA
 MMAE
 AMEM
 EMAM
 MAEM
 MEMA
 MAME
 AMME
Different combinations of letters for the name Lucy:
























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.
Middle
 NHOJ
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 × (N1) × (N2) × (N3) × ... × 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.
 FILIP
 FIPIL
 FILPI
 FIPLI
 FIIPL
 FIILP
 FLIIP
 FLPII
 FLIPI
 FPIIL
 FPLII
 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




















Conclusion
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}.
This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.
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