math
posted by bballer .
Find the number of distinguishable permutations of the given letters ``AAABBBCCC''.
If a permutation is chosen at random, what is the probability that it begins with at least 2 A's?

9!/(1!*3!*3!)
you take the total amount of letters you have factorial, divided by each of you different types of letters factorial. the reason that there is a 1! in the beginning is because you have the first two A's in the front, so they won't be going anywhere, leaving you with 1 A, therefore, 1!. 
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