Vandermonde identity
Theorem 1 ([1] 24.1.1 formula A. II.).
For any p and q and any k with 0≤k≤p+q,
(p+qk)=k∑i=0(pi)(qk-i). | (*) |
Proof.
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 P∪Q of size k.
To build a subset of P∪Q of size k, we first decide how many elements, say i with 0≤i≤k,
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 P∪Q of size k subject to the restriction that exactly i elements come from P. Summing over all possible i completes
the proof.
∎
References
- 1 Abramowitz, M., and I. A. Stegun, eds. Handbook of Mathematical Functions. National Bureau of Standards, Dover, New York, 1974.
Title | Vandermonde identity![]() |
---|---|
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 |