This means that to find the number of arrangements of letters in a word you use this expression:
N!
N! means N factorial.
This means that you times N by all of the numbers below it until you get to 1.
N x (N-1) x (N-2) x (N-3) x (N-4) x (N-5) ………. x 1
Using this expression I am going to work out how many different letter arrangements I think there will be for these names with different numbers of letters and then check them by writing out the arrangements.
TIM
To work out the arrangements for Tim I need to use 3! which is:
3 x 2 x 1 = 6
Having written out all of the arrangements for the letters of the word TIM I can see that I was correct in saying that there would be 6 different arrangements.
TIM TMI MIT MTI IMT ITM
Other lengths of words would therefore be:
2 letters = 2! = 2 x 1 = 2
5 letters = 5! = 5 x 4 x 3 x 2 x 1 = 120
6 letters = 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
Working out the different arrangements for words with repeated letters is different to words with no repeated letters.
I predict that words with 1 letter repeated twice will have fewer arrangements than words with no repeated letters. I think this because for every combination of letters there will be another that is exactly the same because all you have done is swap around the two repeated letters.
Emma
EMMA EMAM EAMM MMAE MAME MEAM
MAEN MEMA MMEA AEMM AMEM AMME EMMA EMAM EAMM MMAE MAME MEAM
MAEN MEMA MMEA AEMM AMEM AMME
The letters in italics have already been written out but using the other m, so therefore there is half the amount of possible combinations. This means that to get the number of possible combinations of a word with one set of letters repeated twice you would have to divide by 2 so that you get half the number of results.
Therefore the expression used to work out the number of possible combinations of a word with one set of letters repeated twice is:
N! This means that you would divide the factorial of
- the length of the word by 2.
I am going to test that my expression is correct by testing it on a 3 letter word with two of the same letters in it.
LEE EEL ELE LEE EEL ELE
This shows that there are 3 possible combinations for the word LEE, I am now going to check my expression by seeing if I get 3 from my expression.
3! = 3 x 2 x 1 = 6
6 = 3
2
This shows that my expression was correct and my prediction was also correct because a word with one letter repeated twice has half the amount of combinations than a word with no repeats.
I am now going to see what would happen if I used a name with 3 repeated letters in it.
BOBB
OBBB BOBB BBOB BBBO
OBBB BOBB BBOB BBBO
OBBB BOBB BBOB BBBO
OBBB BOBB BBOB BBBO
OBBB BOBB BBOB BBBO
OBBB BOBB BBOB BBBO
This shows that the number of combinations for a word with 3 of the same letters is 6 times less than that of a word of equal length with no repeats.
This means that to get the number of different combinations for a word with 3 of the same letter in it you must have to dived the total number of combinations by 6 so that you eliminate those words where the letters are in the same order.
Therefore:
N!
6
I have noticed that both 2 and 6 are the factorials of the number of repeated letters in the word.
For example in the word EMMA there are 2 M’s the factorial of 2 (2 x 1) equals 2
And in the word BOBB there are 3 B’s, the factorial of 3 (3 x 2 x 1) equals 6.
Therefore when a word has any number of one repeated letter the expression you can use is:
N!
R! (when R is the number of repeated letters)
Now that I have found the expression for words with one set of repeated letters I am going to see what happens when you use a word which has more than one set of repeated letters.
I expect that when there is more than one set of repeated letters in a word you will have to divide the factorial of the number of letters in the word by (the number of the first set of repeated letters! X the number of the second set of repeated letters!).
HANNAH
I am first going to try my theory on the name HANNAH
The expression I will use to get the number of combinations is
6! _ _ = 90
2! 2! 2!
I am not going to write out all of the combinations for Hannah because I do not feel that I need to.
PHILLIP
I am now going to test my theory on the name PHILLIP because it has a different number of letters to the word HANNAH.
7! _ _ = 630
2! 2! 2!
Both of these calculations seem to work because they come up with reasonable answers when you compare them to the original N! of the word.
I can see that both of these words work with the expression:
N!
R1! R2!
This expression means that you divide the factorial of the number of letters (N!) by the factorial of the number of first set of repeated letters (R1!) by the factorial of the second set of repeated letters (R2!), this can be done with however many sets of repeated letters there are.
ANASTASIA
I am lastly going to try my expression with the name ANASTASIA
Because it has a lot of letters in the word and two sets of repeated letters:
9! _ _ = 7560
4! 2!
So the expressions for working out the number of combinations available for any word are these:
N! -for a word without any repeats
N! -for a word with one set of repeated letters
R!
N! _ -for a word with any amount of sets of repeated letters
R1! R2!
