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Question

Find how many committees with a chairman can be chosen from a set of n persons. Derive the identity (n1)+2(n2)+3(n3)++n(nn)=n2n1.

A
n(1+x)n1=(n1)+2(n2)x++(n1)(nn1)xn2+n(nn)xn1.
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B
n(1x)n1=(n1)+2(n2)x++(n1)(nn1)xn2+n(nn)xn1.
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C
n(1+x)n+1=(n1)+2(n2)x++(n+1)(nn1)xn2+n(nn)xn1.
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D
n(1x)n1=(n1)+2(n2)x++(n1)(nn+1)xn2+n(nn)xn1.
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Solution

The correct option is A n(1+x)n1=(n1)+2(n2)x++(n1)(nn1)xn2+n(nn)xn1.
The number of persons in the committee may vary between 1 and n. Let us count how many such committees number k persons.The k persons can be chosen in (nk) ways while the chairman can be chosen in k ways, yielding a total of k(nk) committees with k persons.Adding up for k=1, 2, ..., n, we see that the total number of committees is (n1)+2(n2)+3(n3)++n(nn).On the other hand, we can first choose the chairman. This can be done in n ways.Next, we choose the rest of the committee, which is an arbitrary subset of the remainingn-1 person. Because a set with n-1 elements contains 2n1 subsets, itfollows that the committee can be completed in 2n1 ways and the total number of committees is n2n1. Observation There are many proofs of the given equality. An interesting one is the following. Consider the identity (1+x)n=1(n1)x+(n2)x2++(nn1)xn1+(nn)xn. Differentiating both sides with respect to x yields n(1+x)n1=(n1)+2(n2)x++(n1)(nn1)xn2+n(nn)xn1.Setting x=1 gives the desired result.

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