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. You could try this out in the case of 5. When n was 5 and k was 3, this resulted in 5 times 4 times 3. That was this term right over here.