Lucas' Theorem
Lucas' theorem asserts that, for p prime, a ≥ 1 and 0 < k < pa, C(pa, k) = 0 (mod p), where C(n, m) denotes the binomial coefficient "n choose m". Its particular case, where p = 2, was instrumental in establishing a relationship between Pascal's triangle and Sierpinski's gasket. We had to go to a considerable length to prove this variant of the theorem. Surprisingly, there is a short proof based on an engaging property of the binomial coefficients
| (1) |
k·C(n, k) = n·C(n - 1, k - 1).
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This identity admits an elegant combinatorial proof. Assume the task is to choose a k-member committee and its chairman out of a group of n candidates. We can approach the task in two ways:
First select k members of the committee (which can be done in C(n, k) ways) and out of the selected group choose one person to head the committee. This can be done in k ways. In all, there are k·C(n, k) ways to choose a committee and its chairman.
First, choose the committee's chairman out of n candidate. Next choose k - 1 committee members out of the remaining n - 1 candidates. This can be accomplished in n·C(n - 1, k - 1) ways.
(1) merely asserts that the number of ways to compose the committee is the same regardless of the manner in which it is counted.
Now, for Lucas' theorem. Let n = pa, with p prime, a ≥ 1, and 0 < k < n. (1) then becomes
| (2) |
k·C(pa, k) = pa·C(pa - 1, k - 1),
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so that the left hand side has a prime factors p. However, k < pa, which means that k, by itself, may at most be divisible by pa-1, leaving at least one factor p for C(pa, k). In other words, C(pa, k) is divisible by p:
| (3) |
C(pa, k) = 0 (mod p).
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References
- A. T. Benjamin, J. J. Quinn, Proofs That Really Count: The Art of Combinatorial Proof, MAA, 2003
Copyright © 1996-2009 Alexander Bogomolny
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