Choosing colored balls 
There're 5 white balls, 9 black balls, and 14 red balls.  What's the probability that if I pick out 6 balls without replacement, it will be 3 white, 2 black and a red ball?

Help appreciated!
I think the answer is (28 choose 6)/3! 2!, is that correct?
 A: HINT: There are $\binom{28}6$ different ways to pick $6$ balls, and they’re equally likely. If $n$ is the number of ways to pick a combination containing $3$ white balls, $2$ black balls, and one red ball, then the probability of picking such a combination is $$\frac{n}{\binom{28}6}\;:$$ the total number of possibilities goes in the denominator, not the numerator.
To calculate $n$, observe that there are $\binom53$ ways to choose $3$ white balls from the $5$ that are available.


*

*How many ways are there to pick $2$ black balls from a set of $9$?

*How many ways are there to pick one red ball from a set of $14$?

*How should you combine these various numbers to get $n$, the number of successful combinations?
A: This scenario can be best described using the generalized hypergeometric distribution with probability density function 
$$ \mathbb{P}(y_1,...,y_k) =  \frac{\binom{m_1}{y_1}\cdots\binom{m_k}{y_k}}{\binom{m}{n}} $$
where each $y_i\in \mathbb{N}$ , $n=\sum_{i=1}^k y_i$, $m$ is the total number of balls and each $\binom{m_i}{y_i}$ is the total number of unordered sets of size $y_i$ in the whole set consisting only of type $i$ objects.
Looking to your problem we have that the coefficients $\binom{m_i}{y_i}$ are $\binom{5}{3}=10$ ,$\binom{9}{2}=36$ and $\binom{14}{1}=14$. Also $\binom{m}{n}=\binom{28}{6}=376740$. Therefore
$$ \mathbb{P}(3,2,1)=\frac{10\cdot 36 \cdot 1}{376740} \approx 0.000955566$$
