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Question

Prove that if n and r are positive integers
nrn(n1)r+n(n1)2!(n2)rn(n1)(n2)3!(n3)r+
is equal to 0 if r be less than n, and to n! if r=n.

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Solution

We have
(ex1)n=(x+x22!+x33!+x44!+)n
=xn+ terms containing higher powers of x ..... (1)
Again, by the Binomial Theorem,
(ex1)n=enxne(n1)x+n(n1)12e(n2)x ......(2).
By expanding each of the terms enx,e(n1)x, we find that the coefficient of xr in (2) is
nrr!n(n1)rr!+n(n1)2!(n2)rr!n(n1)(n2)3!(n3)rr!+
and by equating the coefficients of xr in (1) and (2) the result follows.

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