Ball Permutations without replacement This is a question from a sample stat midterm. 

A box has 60 balls - 15 are Yellow; 15 are red; 15 are white; 15 are
  green. 14 Balls are selected without replacement and are put in 4
  boxes, one for each ball. 
a) Compute the probability that box 1 has 5 balls, box 2 has 4 balls,
  box 3 has 3 balls, and box 4 has 2 balls.
b) Compute the probability that 2 boxes have 5 balls, 1 box has 3
  balls, and one box has 1 ball.

I was a bit confused on how to approach this. I was thinking that the total number of ways to distribute the balls should be $4^{14}\binom{60}{14}$, so the probability for part a should be something like $\frac{w}{4^{14}\binom{60}{14}}$ where $w$ is the number of ways to fulfill the above requirements. But I'm not sure if that is the correct approach, or where to go from there. What is the correct way to solve this?
 A: I think the problem is unclear.  I expect that what is intended is this: "pick the $14$ balls as described.  What is the probability that there are $5$ of one color, $4$ of a second color, $3$ of a third color, and $2$ of the remaining color?"  
To answer that question (which may well not be the intended one):  Note that there are $4$ ways to choose the first color, and then $\binom {15}5$ ways to choose five of those balls.  Then there are $3$ ways to choose the second color with $\binom {15}4$ ways to choose four of those.  Continuing with this we see that there are $$\binom {15}5\times \binom {15}4\times \binom{15}3 \times \binom {15}2\times 4!$$ combinations that pass the first requirement.  As there are $\binom {60}{14}$ total combinations we see that the probability is given by the ratio:  $$\frac {\binom {15}5\times \binom {15}4\times \binom{15}3 \times \binom {15}2\times 4!}{\binom {60}{14}}$$
A: The colours have nothing to do with the problem. Neither does the fact there were $60$ balls to begin with. 
We will assume that each of the $14$ balls is "thrown" toward the boxes, say one at a time, and that where the various balls land are independent of each other.
It is convenient to assume the balls are distinct (they have ID numbers on them, painted in invisible ink). Then there are $4^{14}$ equally likely possibilities for where the various balls land.
We now count the "favourables." The $5$ balls that land in Box 1 can be chosen in $\binom{14}{5}$ ways. For each of these ways, the $4$ that land in Box 2 can be chosen in $\binom{9}{4}$ ways, and so on.
So the number of favourables is $\binom{14}{5}\binom{9}{4}\binom{5}{3}$. This simplifies to $\frac{14!}{5!4!3!2!}$.
One can get at the same count of favourables using multinomial coefficients. 
A: Hint: Who cares which balls were selected?  That there were originally $60$ balls is a complete multi-colored herring.
We are just looking for the probability that fourteen balls selected are put into the four boxes in the given arrangement. (Assume there's no bias in box assignment.)  
How many ways can we assign boxes to the fourteen balls?  How many ways can box 1 be assigned to 5 balls, box 2 to four balls, box e to three balls, and box 4 to the last two balls remaining?

 $$\frac{14!}{5!~4!~3!~2!}\Big/4^{14}$$

