How do I find probability that a poker hand contains 4 black cards and 2 spades

Question

Suppose 5 cards are drawn from the standard deck of 52 cards. Denote random variables $X$ to be the number of black cards in the hand and $Y$ to be the number of spades in the hand.

How do I calculate $P(X=4,Y=2)$, the probability that a poker hand contains 4 black cards and 2 spades?

My attempt:

$$P(X=4,Y=2)=\frac{{13\choose2}{13\choose2}{26\choose1}}{{52\choose5}}$$

where ${13\choose2}$ is the number of ways to select 2 spades, ${13\choose2}$ is the number of ways to select the remaining 2 black cards excluding spades, and ${26\choose1}$ is the number of ways to choose a red card.

Answer 1

Consider just the spades. There are $13$ of them - you require $2$. There are therefore $13$ ways to pick the first spade, and $12$ ways to pick the second.

However, this involves repeating every pair twice, for example $AK$ and $KA$.

There are therefore $\frac{13\cdot12}{2}=\binom{13}{2}$. Same for clubs.

There are $26$ red cards we only need $1$, so $\binom{26}{1}$.

Now we have our $5$ cards, they can be arranged in the hand in $5!$ ways.

So the total number of hands is $5!\binom{13}{2}\binom{13}{2}\binom{26}{1}$.

There are $52$ cards at first, $51$ next, etc..., until $48$ for the last card. This can be written as $\frac{52!}{(52-5)!}=\frac{52!}{47!}$.

Put the whole lot together therefore gives

$$\frac{\binom{13}{2}\binom{13}{2}\binom{26}{1}}{\binom{52}{5}}$$