Finding the number of ways to pick ${n}$ marbles from a jar Problem: 
А jar contains 8 blue marbles, 6 green marbles, and 4 red marbles. Five marbles are selected at random, all at once. In how many ways can: 
A.) two red and three blue marbles be obtained?
B.) two green and two blue marbles be obtained?
C.) two red marbles be obtained?
D.) two or more green marbles be obtained?

Work:
A.) $4\cdot3\cdot8\cdot7\cdot6 $ $\cdot$ $_{5}C_{2}$ = $40,320$
B.) $6\cdot5\cdot8\cdot7\cdot6 $ $\cdot$ $_{5}C_{2}$ = $100,800$
C.) $4\cdot3\cdot12\cdot11\cdot10 $ $\cdot$ $_{5}C_{2}$ = $158,400$
D.) ($6\cdot5\cdot12\cdot11\cdot10 $ $\cdot$ $_{5}C_{2}$) + ($6\cdot5\cdot4\cdot12\cdot11 $ $\cdot$ $_{5}C_{3}$) + ($6\cdot5\cdot4\cdot3\cdot12 $ $\cdot$ $_{5}C_{4}$) + ($6\cdot5\cdot4\cdot3\cdot2 $ $\cdot$ $_{5}C_{5}$) = $576,720$
Is all of work correct?

We assume (as we are expected to) that the marbles are distinguishable.
A.) There are $\binom{4}{2}$ ways to choose the reds. For each of these ways, there are $\binom{8}{3}$ ways to choose the blues, for a total of $\binom{4}{2}\binom{8}{3}$.
B.) We interpret the question as asking for exactly $2$ green and exactly $2$ blue, so $1$ red. The same reasoning as above gives $\binom{6}{2}\binom{8}{2}\binom{4}{1}$. 
C.) The $2$ red can be chosen in $\binom{4}{2}$ ways. Now we must choose $3$ non-red from the $14$ non-red. 
D.) It's your turn.
