This is going to be equal to 2 to the nth possibilities. Let's think about how many of those equally likely possibilities involve k heads. Just like we did for the case where we're thinking about 3 heads, we say, well look, the first of those k heads, how many different buckets could it fall into? 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.