If you have an urn with $4$ balls each of red, white, blue and black, what is the probability distribution of getting drawing $4$ balls and getting exactly $0$, $1$, $2$, $3$, or $4$ red balls?
My attempts to calculate combinations for drawing exactly $1$ red balls is as follows: $$\binom{4}{1}\binom{12}{3}$$ where $\binom{4}{1}$ counts the number of ways of selecting one red balls and $\binom{12}{3}$ counts the number of ways of selecting three of the other twelve balls, with a result of $4\times220=880$.
I figure the total combinations would be $4^4$ or $256$, so I'm obviously on the wrong track.
As lulu pointed out in the comments, the number of ways four of the sixteen balls can be selected is $$\binom{16}{4}$$
The number of ways of selecting exactly $k$ red balls when four balls are drawn from the urn is $$\binom{4}{k}\binom{12}{4 - k}$$ where $\binom{4}{k}$ is the number of ways of selecting $k$ of the four red balls and $\binom{12}{4 - k}$ is the number of ways of selecting $4 - k$ of the other $12$ balls in the urn.
Hence, the probability of selecting exactly $k$ red balls when four balls are drawn from the urn is $$\frac{\dbinom{4}{k}\dbinom{12}{4 - k}}{\dbinom{16}{4}}$$ where $0 \leq k \leq 4$.