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In Generatingfunctionology (2nd ed available free here https://www.math.upenn.edu/~wilf/DownldGF.html ), nCk for integer n < 0 is defined as

nCk = n*(n-1)...(n-k+1)/k!

Then he asserts that:

sum (nCk)*y^n with n ranging over all integers, k is an integer >= 0, summand vanishes whenever n < k

I see why the summand would vanish for nonnegative integer n < k, because n*(n-1)...(n-n)...(n-k+1) = 0, but why for negative integer n? The text hasn't been corrected in the third edition, so I guess it's not an error? Or is it because nCk = nC(n-k), so if n is negative and k nonnegative n - k is negative, and that makes nC(n-k) = 0? Then how is that consistent with the earlier claim that nCk is nonzero for negative n and nonnegative k?

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This appears to be a typo; $\sum_{n} {n\choose k} y^n$ should be $\sum_{n\ge 0} {n\choose k} y^n$.

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  • $\begingroup$ The OP wonders how to reconcile the two statements from the book, and I have provided an explanation which I believe correct. $\endgroup$ – vadim123 May 12 '18 at 17:05

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