The first of the k heads could fall into n different buckets. It could be the first flip, second flip, all the way to the nth flip. Then the second of those k heads, we don't know exactly how many k is, will have n minus 1 possibilities. The third of those k heads will have n minus 2 possibilities, since two of the spots are already taken up. We would do this until we have essentially accounted for all of the k heads. This will go down all the way to, we will multiply the number of things we're multiplying is going to be k. One for each of the k heads. This is 1, 2, 3, and then you're going to go all the way to n minus k minus 1.