combinatorics, you can win prizes from different categories but not from the same one. The question goes like this:

There is a conference for psychology in which $12$ researchers are participating. In the conference, two different companies are giving out prizes in two different categories. Three researchers will receive an award for "the most innovative research" (TV, DVD, or a radio), while two researchers will receive an award "the best presentation" (cash prize overall 10,000 dollars). If you know that every researcher can win in more than one category but can't win more than one prize in each category, how many different ways are there to divide the prizes?

choose one:
A) $14520$
B) $87120$
C) $95040$
D) $174240$
What I did was say each one of the researchers can win one prize of the first category, that's $C(12,3)$, then each one of the researchers can also win the second category, which is $C(12,2)$, which got me to $14520$, but apparently the answer is $87120$.
 A: The first three prizes are distinct (TV, DVD, radio), so we use permutation instead of combinaison. There are $P(12,3)=1320$ ways to give the first three prizes.
The second prizes were the same, si combinaison was right here. There are $C(12,2)=66$ ways the give the second prizes.
Finally, there are $P(12,3)\times C(12,2)=87120$ wyas to give all the prizes.
A: You overlooked the fact that there are three different prizes for "the most innovative research" prize.  Also, we have to account for the number of people who win both prizes.
Nobody wins both prizes:  There are $\binom{12}{3}$ ways to select the winners of "the most innovative research" prize and $3!$ ways to distribute those prizes to the winners.  There are $\binom{9}{2}$ ways to select the winners of "the best presentation" prize to the remaining people.  Hence, there are
$$\binom{12}{3}3!\binom{9}{2}$$
ways to distribute the prizes if nobody wins both prizes.
Exactly one person wins both prizes:  There are $12$ ways to select the recipient of both prizes.  There are $\binom{11}{2}$ ways to select the other two winners of "the most innovative research" prize, $3!$ ways to distribute the prizes for "the most innovative research" prize among the winners, and $9$ ways to select the other winner of "the best presentation" prize.  Hence, there are
$$\binom{12}{1}\binom{11}{2}3!\binom{9}{1}$$
ways to distribute the prizes if exactly one person wins both prizes.
Two people win both prizes:  There are $\binom{12}{2}$ ways to select the two people who win both prizes, $10$ ways to select the other winner of "the most innovative research" prize, and $3!$ ways to distribute the prizes for "the most innovative research" prize among the winners.  Hence, there are 
$$\binom{12}{2}\binom{10}{1}3!$$
ways to distribute the prizes if two people win both prizes.
Since the three cases are mutually exclusive and exhaustive, the number of ways the prizes can be distributed is 
$$\binom{12}{3}3!\binom{9}{2} + \binom{12}{1}\binom{11}{2}3!\binom{9}{1} + \binom{12}{2}\binom{10}{1}3!$$
