AAAA has no variations.
3 letters
The name LOU has three letters, all different.
LOU ULO OUL
LUO UOL OLU
The name LEE has 3 letters, 2 the same.
LEE EEL ELE
The name CCC has 3 letters, all the same.
CCC has no variations.
2 letters
The name LI has 2 letters, both different.
LI IL
The name DD has 2 letters, both the same.
DD has no variations.
After carrying out these experiments, I have come up with the following table of results:
From this table I see a pattern. In 2 letters, 1 is ½ of 2. In 3 letters, 1 is of
3 which is ½ of 6. In 4 letters, 4 is Of 12 which is ½ of 24. This means that I can predict what 5 will be. I can do this because of the pattern I noticed.
2 X 3=6 which is the number of combinations a 3 letter word can have.
6 X 4=24 and with 4 letters the number of combinations is 24. I shall multiply 5 X 24 to get the combinations for 5 letters. I will then half the answer I get from this to find out the combinations for 5 letters with 2 the same. I’ll divide this answer to find the combination for a 5 letter word with 3 the same, and so on.
From these results I have now found out how this works. The number of combinations is narrowed down each time there is more that 1 letter the same, or 1 letter is chosen to begin.
This is what I mean:
E.g. 2 letter word
2 choices for letter, multiplied by one choice for letter = 2
3 letter word
1st letter 2nd letter 3rd letter
3 X 2 X 1 = 6
4 letter word
1st letter 2nd letter 3rd letter 4th letter
4 X 3 X 2 X 1 = 24
And I predict, for a 5 letter word…
1st letter 2nd letter 3rd letter 4th letter 5th letter
5 X 4 X 3 X 2 X 1 = 120
From this, we can see why, in Emma’s case, the choice is halved once the first letter is chosen. 2 letters are the same the combinations are halved. E.g. If we use E as the 1st letter, we only have M or A as our second choice.
So if 3 letters are the same, it is of 2 letters the same which equals of ½ of all different letters - in total, Of all different letters.
And, if 4 letters are the same, its ¼ of 3 letters the same I.e. Which is Of all letters being different.
An easier method for doing this it to use the factorial button on my scientific calculator : [ ! ] will multiply all the numbers below a certain number, which is what I can do to predict the amount of combinations for the letter 5.
E.g. 10 ! = 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1
10! = 3, 628, 800
I can therefore say that a 10 letter word with all of the letters different could make 1, 628, 800 combinations of itself.
I have come up for this equation for finding the combination of an n letter word, with all letters different.
N! = n x (n - 1) x (n - 2) x (n - 3) x (n - 4) x (n - 5) x (n - 6) x (n - 7) … etc.
If 2 letters were the same, the equation would be:
n! 2
And if 3 letters were the same, it would be:
n! 2 3 …etc.
This concludes my coursework, as I have now found how many combinations you can get from an n letter word, even when the word has 2 letters the same, 3 letters the same, 4, or 5 …etc.
- Louise Manly
