# Homework Help: A probability question

1. Sep 23, 2008

### kingwinner

1) A company orders supplies from M distributors and wishes to place n orders (n<M). Assume that the company places the orders in a manner that allows every distributor an equal chance of obtaining any one order and there is no restriction on the number of orders that can be place with any distributor. Find the probability that a particaular distributor--say, distributor I--gets exactly exactly k orders (k<n).

This is an example in my textbook and the answer given is probability=[nCk * (M-1)^(n-k)] / M^n. I don't understand why this is correct. (M-1)^(n-k) and M^n assume the order is important (some sort of "permutation with replacement"), but nCk is a combination in which order is not important. Why is there combination at the top and permutation at the bottom?
What I also don't understand is why wouldn't the answer be [nPk * (M-1)^(n-k)] / M^n? This way we would have permutation divided by permutation.

Thanks for explaining! (I think no one would be able to explain this without a very firm understanding or probability, but I am sure someone here can)

Last edited: Sep 23, 2008