An algebra problem by Aritra Chakraborty

50c25 can be written as ${2}^{3}{3}^{2}{37}^{1}{7}^{2}{41}^{1}{43}^{1}{13}^{1}{47}^{1}{29}^{1}{31}^{1}$ . 18 can be written as ${2}^{1}{3}^{2}$ .

From this it's clear that the limiting factor (pun intended) is the number of 9's available to make 18's. There's just one of them.

Edit: I knew there must also be a much more clever way of doing this. There is. Here's one explanation of the algorithm: nCr%M