For example, for $4$ men $\{M_1...M_4\}$ and $3$ Women $\{W_1...W_3\}$, how many potential matches are there if the men go one by one proposing so one man ends up single? I know the answer is $73$ but cannot figure out how to get $73$.
2026-03-25 12:54:30.1774443270
Combinatorics - how many potential matches between n men and m women where n>m?
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In order to justify the answer $73$ one must re-write the problem and allow more than one man to stay single. With this assumption, then let $C_i$ be the number of matches in which exactly $i$ men stay single. Then, clearly, $$C_i=\binom 4i\times 3\times \cdots \times (3-(i-1))$$ We quickly see that the answer is then $$\sum_{i=1}^4 C_i=1+12+36+24=73$$
A similar calculation works for the general problem. I don't know if there is a simple closed representation for the sum.