Tricky subset counting problem

Question

I had a combinatorics test today and I came upon this question which I just couldn't figure out how to do. Can anyone explain in depth how to do this?

Let $m \ge 34$ be an even integer,
Let $n \ge 1$ be an integer.

Consider the two sets

$$ \begin{gather} \begin{aligned} A &= \{1,2,\ldots,m\} \\ B &= \{m+1,m+2,\ldots,m+n\} \end{aligned} \end{gather} $$

Let $k$ be an integer with $17 \le k \le n+17$.

Consider the subsets $X$ of $A \cup B$, such that $|X| = k$, $|X \cap A| = 17$, and all elements of $X \cap A$ are even.

How many such subsets are there?

The answer is: ${{m/2}\choose17} \cdot {{n}\choose{k-17}} $

I am having trouble figuring out what this means.

Consider the subsets $X$ of $A \cup B$, such that $|X| = k$, $|X \cap A| = 17$, and all elements of $X \cap A$ are even.

I believe I will better understand this problem if I understand how to break down these set notations and get a more intuitive view of this problem which is easier to grasp.

Thank you!

Answer 1

Consider the subsets $X$ of $A \cup B$, such that $|X| = k$, $|X \cap A| = 17$, and all elements of $X \cap A$ are even.

What this means in English is, the subset $X$ has $k$ elements and $17$ of these should be even elements from the set $A$. That's where ${m/2}\choose17$ comes from.

And for the rest of the elements ($k-17$ left), we should choose them from $B$ because we have already chosen $17$ elements from $A$ and the condition $|X \cap A| = 17$ is satisfied. So we can choose them with ${n}\choose{k-17}$. Then the result follows.

Answer 2

Note that the "first" $17$ elements of $X$ must be chosen from $A$ and must be even, giving $ \binom{m/2}{17}$ways. Now we need to choose $k-17$ elements from $B$ ($B$ has cardinality $n$) so that is $\binom{n}{k-17}$. And we have $\color{red}{\binom{m/2}{17} \binom{n}{k-17}}$ in total.