Consider the pairs of opposite faces of the cube. Observe that every move increases the sum of every pair by 1. Thus, if the cube eventually has the same number on all its faces, each pair must have the same value at the start. The sum of all the faces is 60, so the sum of each pair must be 20. Furthermore, if the sum of each pair is 20, then the numbers on the cube can be made all equal: Note that if you pick one face from each pair, the resulting three faces will always share a common vertex. So, let $x$ be the maximum integer on the cube. Simply increase each face until it hits $x$. Since the face pairs all have equal sum, the number of increases needed by each face pair is equal, so you can combine the increases into valid moves. Thus, the task is possible iff $M$ can be partitioned into three pairs, the sum of each pair of which is 20. There are 9 pairs of positive integers that sum to 20: $\{x, 20-x\}$ for $1 \leq x \leq 9$, and we may pick any 3 of them to form $M$, so the answer is $\binom{9}{3} = \boxed{84}$.