Ways To Count
The purpose of this page is to present several proofs of an identity that involves binomial coefficients:
| (1) |
C(n, 1) + 2·C(n, 2) + 3·C(n, 3) + ... + n·C(n, n) = n2n-1.
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Four such proofs have been collected in a 1999 issue of Crux Mathematicorum by Jimmi Chui, then a secondary school student. I found two more elsewhere and present all six below.
Pure Algebra
Let S denotes the sum in (1). Using the basic property of the binomial coefficients C(n, k) = C(n, n - k) we have
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| S | = 0·C(n, 0) + 1·C(n, 1) + 2·C(n, 2) + ... + n·C(n, n) |
| | = 0·C(n, n) + 1·C(n, n-1) + 2·C(n, n-2) + ... + n·C(n, 0) |
| | = n·C(n, 0) + (n-1)·C(n, 1) + (n-2)·C(n, 2) + ... + 0·C(n, n). |
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Adding the first and the last sums gives
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| 2S | = n·C(n, 0) + n·C(n, 1) + n·C(n, 2) + n·C(n, 3) + ... + n·C(n, n) |
| | = n·2n, |
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which immediately applies (1).
Mathematical Induction
We shall prove (1) by mathematical induction. For n = 1, there is nothing to prove. Assume (1) holds for some n = k:
| (2) |
C(k, 1) + 2·C(k, 2) + 3·C(k, 3) + ... + k·C(k, k) = k2k-1.
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If the left side (2) is denoted S(k), we have using the addition formula
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| S(k+1) | = C(k+1, 1) + 2·C(k+1, 2) + 3·C(k+1, 3) + ... + (k+1)·C(k+1, k+1) |
| | = 1·(C(k, 0) + C(k, 1)) + 2·(C(k, 1) + C(k, 2)) + ... + k·(C(k, k-1) + C(k, k)) + (k+1)C(k, k) |
| | = 1·C(k, 0) + 3·C(k, 1) + 5·C(k, 2) + ... + (2k+1)C(k, k) |
| | = (C(k, 0) + C(k, 1) + C(k, 2) + ... + C(k, k)) + 2·S(k) |
| | = 2k + 2·k·2k-1 |
| | = (k + 1)·2k. |
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Generating functions
Consider the generating function for the binomial coefficients:
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f(x) = C(n, 0) + C(n, 1)·x + C(n, 2)·x2 + ... + C(n, n)·xn
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The derivative of which is
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f '(x) = C(n, 1) + 2·C(n, 2)·x1 + ... + n·C(n, n)·xn-1.
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So that f '(1) = S(n). On the other hand, by the binomial theorem,
with the derivative
from which f '(1) = n2n-1.
Combinatorial Proof 1
Consider a set of n people. We shall answer the question, "In how many ways it is possible to select a (non-empty) committee and its chairperson from the set of n people?" We shall answer the question in two ways.
There are C(n, k) committees with k members (k ≥ 1) and, for each, there are k choices of a chairperson. Thus S(n) is the total number of such committees with a chair.
We may also start by selecting the chairperson first and then complementing the committee by regular members out of the remaining n-1 fellows. There are n choices for the chairperson and 2n-1 possibilities for other members, which gives the right hand side in (1).
Combinatorial Proof 2
The expression S(n)/2n [Benjamin & Quinn, p. 66] also has a combinatorial interpretation. It answers the question, "What is the average size of a subset of {1, 2, ..., n}?" Indeed, there are exactly C(n, k) subsets with k terms whereas 2n is the total number of the subsets.
On the other hand, pair each subset with its complement. The two together have n terms and the average of n/2. So that
which is equivalent to (1).
Formula for Binomial Coefficients
Recollect that [Zeitz, p. 215]
Then
where the sum is taken from k = 1 to k = n. Thus
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| S(n) | = ∑ |
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| | = ∑ |
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| | = ∑ |
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| | = ∑ |
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| | = n·∑ |
| (n-1)! |
| (k-1)!((n-1)-(k-1))!, |
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Applying the binomial theorem.
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| S(n) | = n·∑C(n-1, k-1) |
| | = n·(C(n-1, 0) + C(n-1, 1) + ... + C(n-1, n-1)) |
| | = n·2n-1, |
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References
- A. T. Benjamin, J. J. Quinn, Proofs That Really Count: The Art of Combinatorial Proof, MAA, 2003
- P. Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999
Copyright © 1996-2008 Alexander Bogomolny
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