Right, or combinations.

C(50,3) = 50!/[(47!)(3!)] = 50*49*48/3! That is, there are 50 ways to choose the first, 49 ways to choose the 2nd, and 48 ways to choose the 3rd, then we divide by 3! = 6 since there are 3! ways to get the same 3 in a different order. If we didn't divide by 3! we would have

P(50,3) or "permutations of 50 things taken 3 at a time".