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!
It's combinations of orders, since each combination will have many permutations, all equivalent...? That's usually why you use combinations not permutations, to avoid redundant cases. In this case, an order for a and b, is equivalent to an order for b and a, from a supplier. (Just guessing)
I don't quite get your point. But most definitely M x M x ... x M = M^n is assuming that order matters (permutation with replacement), and nCk at the top is a combination. I still don't understand why this inconsistency can possibly lead to a correct calculation of the probability. Could someone explain further? Thanks!
Let's try to think of a similar problem.. Let's say that there are 10 companies and 7 orders and you want to find the probability of a company getting exactly 4 orders. Imagine 10 boxes and 7 balls numbered 1 thru 7. Now what is the probability of getting exactly 4 balls in one specific box? Well, for each ball you have 10 choices and 7 balls so the total numbers of ways is 10^7. So we have the denominator. Now the 4 balls can be any 4 of the 7 so we have 7C4. The 3 remaining can go in any of the 9 boxes remaining so we have 9^3. For each of these possibilities we can select any 4 for the box. So we have (7C4 * 9^3)/10^7. Hopefully this helps.
