# Perfoming permutation count

I have a question, probably relatively simple one. Once you want to find all the unique combinations of number 5 you count 5! = 120. Once you want to find all the combinations including repetition of value of number 5 you count 5 power of 5 = 3125.

In my case I have the following situation. I have 9 characters you need to use to perform unique permutations on 4 columns/variables, meaning, for example, having 'a, b, c, d, e, f, g, h, i' I perform all the possible unique 4 combinations by using the 9 characters: 'a b c d'; 'a b e f'; 'd e f g' etc. In each column of each string of 4 characters (e.g. 'a, a, b, c' not allowed) the value of character can't get repeated. Now, how to count the number of such unique combinations obeying the constrains? Cheers

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Hmmm...you said "using all the $9$ characters". By that, do you mean that each character must be used at least once? If so, I'll need to change my answer.... – Cameron Buie Sep 5 '12 at 14:41
-1: It's difficult to tell what's being asked here. – Austin Mohr Sep 5 '12 at 14:43
You have 9 characters and you need to perform all the possible permutations in 4 columns (e.g. 'a, b, d, e') such that you never get repetition of the same character in a single string of the 4 column characters. – ucas Sep 5 '12 at 14:48
With 4 characters a, b, c, d and 2 "columns", would you expect to obtain a count of 6 (because there are ab, ac, ad, bc, bd, cd)? I just wonder - you seem to disallow repetition, but seem to implicitly order. If this is what you want to count, the answer is $n\choose k$, so e.g. ${9\choose 4} = 126$. – Hagen von Eitzen Sep 5 '12 at 14:56