A Mixed Committee Count

Question

A desk has 5 traders and 4 quants. You must pick a 3-person committee. Counting two ways, evaluate sum_k=0^3 binom5kbinom43-k, and say which standard identity this is.

Accepted Answer

Count the same committees two ways. A 3-person committee is just a choice of 3 from the 5 + 4 = 9 people, so directly: binom93 = 84. Now group by the number k of traders on the committee. With k traders and 3-k quants, there are binom5kbinom43-k such committees, and summing over the disjoint cases k = 0, 1, 2, 3 recounts every committee exactly once: binom50binom43 + binom51binom42 + binom52binom41 + binom53binom40 = 4 + 30 + 40 + 10 = 84. This is Vandermonde's identity, sum_k binom mkbinom nr-k = binomm+nr. The proof technique, partition one set of objects by a feature and sum the pieces, is the single most reusable identity-proving move in counting. The trap: misaligning the second index, writing binom4k instead of binom43-k, which silently counts committees of the wrong size. Always check the indices sum to the committee size: k + (3-k) = 3. Interviewers use this to see whether you reach for a combinatorial argument rather than grinding the algebra.

A Mixed Committee Count

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