I would like to ask if you know of any combinatorial (double counting) arguments for finding a closed formula for the following sums:

  1. $\sum_{k=m}^n{k\choose m}{n\choose k}$, where ${n\choose k} = 0$ for $n < k$.
  2. $\sum_{k=1}^n {k\choose m}\frac{1}{k}$.
  3. $\sum_{k=0}^n {k\choose m}k$.

Thank you for any ideas.








Now apply hockeystick identity to find closed forms for 2) and 3).

  • $\begingroup$ Yes, but can you give a combinatorial argument? In a particular a double counting arugment, where you show that both sides of the identity count the same thing? $\endgroup$ – pizet Nov 1 '18 at 10:14

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