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A weighted sum of binomials

HardCombinatoricsBrainteasers~9m

Evaluate 12(33)+11(43)+10(53)+⋯+2(133)+(143)\displaystyle 12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3} + \cdots + 2\binom{13}{3} + \binom{14}{3} (the coefficients count down 12,11,…,112, 11, \dots, 1 as the binomial's top index runs 3,4,…,143, 4, \dots, 14).

Show hints (2)+
  1. The sum is ∑n=314(15−n)(n3)\sum_{n=3}^{14}(15-n)\binom{n}{3}. The descending weight lets you write it as nested hockey-stick sums.
  2. Apply ∑n(n3)=(N+14)\sum_{n}\binom{n}{3}=\binom{N+1}{4} twice; you land on (165)\binom{16}{5}.

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Asked at: Jane Street, Two Sigma

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