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Given two sets (not neccesarily of same size), how many ways can we choose two subsets of same size, each from a different set? Order here doesn't matter and empty subsets count too, also I don't need the answer which includes binomial coefficients. I tried thinking about the Cartesian product of both sets' partitions and how we can eliminate pairs with different lengths but couldn't figure it out.

Example: given two sets {0,1} and {2,3,5} there are ${2\choose 0}*{3\choose 0}+{2\choose 1}*{3\choose 1}+{2\choose 2}*{3\choose 2}$ possible combinations

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  • $\begingroup$ What do you mean by the length of a subset? Its size? So if A has six members and B has eight, there are ${6 \choose 2}{8 \choose 2}$ ways to choose two elements of A and two of B? $\endgroup$ – Ross Millikan Nov 5 '18 at 17:38
  • $\begingroup$ yeah, I meant size. And no that's not what I meant I'll try to make my question a bit more clear $\endgroup$ – QApps Nov 5 '18 at 17:43
  • $\begingroup$ I added an example, I hope it's clear now $\endgroup$ – QApps Nov 5 '18 at 17:53
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Suppose $|A| = n, |B| = m$ (and that $n \leq m$). We want to pick a two subsets of size $k$, one from $A$ and one from $B$. Note that we can do this in $\binom{n}{k} \binom{m}{k}$ ways; summing over all $k$ gives $\sum\limits_{k=0}^n \binom{n}{k} \binom{m}{k}$.

But it turns out that there is a closed-form expression for this. Note that $\binom{n}{k} = \binom{n}{n-k}$. So our sum is equal to $\sum\limits_{k=0}^n \binom{n}{n-k} \binom{m}{k}$. In other words, how many ways can we choose $k$ elements of $B$ and $n-k$ elements of $A$ as $k$ ranges over all possible values? But this merely asks how many ways are there to pick $n$ elements from $A \cup B$ (assuming the sets are disjoint), which is $\binom{m+n}{n}$. Intuitively, this counts the number of ways to choose $n-k$ elements of $A$ to exclude and $k$ elements of $B$ to include.

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  • $\begingroup$ Thanks! This was a very helpful answer $\endgroup$ – QApps Nov 5 '18 at 18:07

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