A weighted sum of binomials

HardCombinatorics Brainteasers~9m

Evaluate $\displaystyle 12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3} + \cdots + 2\binom{13}{3} + \binom{14}{3}$ (the coefficients count down $12, 11, \dots, 1$ as the binomial's top index runs $3, 4, \dots, 14$ ).

Show hints (2)+

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

Answer

Reveal answer →

4368

Create a free account to see the full worked solution and track your progress.

Create a free account

Interview format: Market Maker, Systematic & Data-Driven Trading

A weighted sum of binomials

Answer

Related questions