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?

Question

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.

Answer 1

As indicated in the comments, your answer is correct.

We can select $20$ of the $50$ people for group $A$ in $\binom{50}{20}$ ways, $18$ of the remaining $30$ people for group $B$ in $\binom{30}{18}$, and all $12$ of the remaining $12$ people for group $C$ in $\binom{12}{12}$ ways. Hence, the number of ways of selecting $20$ people for group $A$, $18$ people for group $B$, and $12$ people for group $C$ is $$\binom{50}{20}\binom{30}{18}\binom{12}{12} = \frac{50!}{20!30!} \cdot \frac{30!}{18!12!} \cdot \frac{12!}{12!0!} = \frac{50!}{20!18!12!0!} = \frac{50!}{20!18!12!}$$ as you found.

Answer 2

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 $\frac{50!}{20!18!12!}$.