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Vandermonde Identity: Splits a combination across two groups: C(m+n, r) = sum of C(m,k)C(n,r-k). This concept is essential for quantitative trading interviews and is frequently tested at top firms.

By Valenke Exam Prep Team·Last updated 2026-06-01
Combinatorics

Vandermonde Identity

Splits a combination across two groups: C(m+n, r) = sum of C(m,k)C(n,r-k).

The Vandermonde identity (or Vandermonde convolution): (m+nr)=k=0r(mk)(nrk)\binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} Intuition: Choose rr items from a pool of mm red and nn blue objects. You can take kk red and rkr-k blue for each valid kk. When to use: Problems where you're choosing from two distinct groups, or when you see a sum of products of binomial coefficients that you need to simplify. Special case: m=n=rm = n = r gives (2nn)=k=0n(nk)2\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}^2. Alternative approach: Double counting or a combinatorial argument (the "committee from two clubs" story) derives this from first principles without memorizing the formula.

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