Combinatorial argument for this problem:

Question

Problem: find : $\sum_{r=0}^{21}\binom{42+r}{42}*\binom{42-r}{21}$ ,is there a combinatorial argument for this summation? My progress i realized that sum of upper terms in binomial is constant equal to 84 : so i thought of somehow distributing 84 items in such a manner that 42 gets some and 21 are given with some restrictions but that doesnt work

Answer 1

Perform a change of index $j=r+43$ to rewrite your sum as $$\sum_{j=43}^{64}\binom{j-1}{42}\binom{85-j}{64-43},$$ which you can combinatorially prove is $$\binom{85}{64}$$ by counting the $64$-subsets of $\{1,\dots,85\}$ according to the $43$rd smallest element.