How many ways are there for 50 people to divide them into three groups, A, B and C such that each consists of 20, 18, and 12, respectively? I have tried to solve this problem but I can not figure out where to start. Any help would be appreciated.
Thanks,
EDIT: After another attempt I am leaning towards the answer $\frac{50!}{20!18!12!}$
EDIT2: I know that the answer that I arrived at is correct but I still do not understand why it is correct.

We can split them this way:
We line everyone up in a line (we will get $50!$ this way), then we choose the first 20 into $A$, the next 18 into $B$, and the last 12 into $C$.
Then inside $A, B, C$ we should remove the 'duplicates', i.e. the number of permutations inside the group, which is $20!,18!,12!$ respectively for $A, B, C$.
Hence we arrive at your answer $\frac{50!}{20!18!12!}$.
