A local fraternity is conducting a raffle where $55$ tickets are to be sold - one per customer. There are $3$ prizes to be awarded. If the $4$ organizers of the raffle each buy one ticket, what are the follow probabilities?
(a) The probability that the $4$ organizers win all of the prizes?
For part (a), I was told the answer was $\dfrac{ \binom{3}{3} \binom{52}{1}}{\binom{55}{4}}$
But why wouldn't it be $\dfrac{ \binom{4}{3} \binom{51}{1}}{\binom{55}{4}}$ considering there are $4$ organizers?
(b) The probability that the four organizers win exactly two of the prizes?
For part (b) I need to understand how to do part (a) correctly. So I'm guessing it would either be $$\dfrac{ \binom{3}{2} \binom{52}{2}}{\binom{55}{4}} \text{ or } \dfrac{ \binom{4}{2} \binom{51}{2}}{\binom{55}{4}}$$
The denominator in (a) is total number of unordered 4-plets - all combinations of tickets organizers can get if we don't care which organizer got what. So numerator should also be number of unordered 4-plets consisting of 3 winning tickets and 1 losing. And number of such 4-plets is equal to number of losing tickets.