There are no elements that are in both $ A $ and $ B $.
Because $ | X \cap A | = 17 $, we know 17 elements are from $ A $.
Because all elements of $ X \cap A $ are even, we know 17 of the elements from $ A $ are all even.
Because of this, we can only use half of the elements from $ A $ to fill the 17 even elements: $ \frac{m}{2} $
We choose 17 elements from half of $ A $: $ \binom{m/2}{17} $
The remaining $ k-17 $ elements must come from $ B $: $ \binom{n}{k-17} $
Thus, there are $ \binom{m/2}{17} \cdot \binom{n}{k-17} $ such subsets.