1. C(7,6) = 7!/6!1! = 7. Or: Equivalently, Mary can pick one day to NOT study, and then study on every other day. There are 7 ways to choose one day out of the 7 days of the week. 2. m is larger than n. The operation of erasing the labels on the 70 balls collapses several of the m configurations of labelled balls into one of the n configurations of identical balls. Every configuration of identical balls can be obtained in this way, so erasing labels can only decrease the number of configurations. Or: Given a configuration of identical balls, I can find at least one way to paint labels onto the balls using each number from 1 to 70 exactly once -- and usually there are many ways to do it. No two different configurations of identical balls can ever be painted to end up with the same configuration of labelled balls, so there are more ways to distribute the labelled balls. Or: Let T = the set of ways to distribute 70 identical balls among 2008 bins, and let S = the set of ways to distribute 70 labelled balls among 2008 bins. Define the function f:S->T as follows: given any configuration in S, we erase the labels and get a configuration in T. f is surjective (for every t in T, there exists s in S with f(s)=t), so |T| <= |S|, i.e., n <= m. In fact, it is easy to find a pair of configurations s,s' such that f(s)=f(s'), so we must have n < m. (In fact, m is much, much larger than n.)