Vandermonde identity

Theorem 1 ([1] 24.1.1 formula A. II.).

For any p and q and any k with 0kp+q,

(p+qk)=i=0k(pi)(qk-i). (*)

Let P and Q be disjoint sets with |P|=p and |Q|=q. Then the left-hand side of Equation (*) is equal to the number of subsets of PQ of size k. To build a subset of PQ of size k, we first decide how many elements, say i with 0ik, we will select from P. We can then select those elements in (pi) ways. Once we have done so, we must select the remaining k-i elements from Q, which we can do in (qk-i) ways. Thus there are (pi)(qk-i) ways to select a subset of PQ of size k subject to the restrictionPlanetmathPlanetmathPlanetmathPlanetmath that exactly i elements come from P. Summing over all possible i completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof. ∎


  • 1 Abramowitz, M., and I. A. Stegun, eds. Handbook of Mathematical Functions. National Bureau of Standards, Dover, New York, 1974.
Title Vandermonde identityMathworldPlanetmath
Canonical name VandermondeIdentity
Date of creation 2013-03-22 14:08:49
Last modified on 2013-03-22 14:08:49
Owner mps (409)
Last modified by mps (409)
Numerical id 8
Author mps (409)
Entry type Theorem
Classification msc 05A19
Related topic PascalsRule