Numerically, say I have two numbers: 00000 and 11000
Then I find out each of their HD=2 values:
For 00000 the list is:
5C2 possibilities
For 11000 the list is:
5C2 possibilities
My questions is how do I know the number of intersected binary numbers?
In this case, they are 6 [01001, 01010, 01100, ... 10100]
Form the XOR of the two numbers. There can only be intersections if it has $0$, $2$ or $4$ ones. If it has $0$ ones, we need to modify both strings at the same two arbitrary bits, so the number of intersections is $\binom n2$, where $n$ is the length of the strings. If it has $4$ ones, we can choose $2$ bits at which to modify the one string and then modify the other string at the other two, so there are $\binom42=6$ intersections. If it has $2$ ones (as in your case), we can choose one of the two at which to modify the one string, modify the other string at the other, and then choose one more bit at which to modify both, so here we have $2(n-2)$ intersections (which in your case also comes out to $6$).