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$f(m,n)=\binom{m}{m}\binom{n}{1}+\binom{m}{m-1}\binom{n}{2}+\cdots+\binom{m}{2}\binom{n}{m-1}+\binom{m}{1}\binom{n}{m}=?$

Can we simplify the expression above?

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This type of expression is very well known. See Concrete Mathematics by Graham, Knuth, and Patashnik, particularly table 174 (p. 174). –  Todd Trimble Oct 2 '10 at 0:21
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(I should have paused; this looks like homework.) –  Todd Trimble Oct 2 '10 at 0:23
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closed as too localized by Ben Webster Oct 2 '10 at 3:51

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