$$\sum_{k=i}^{n} \binom{n}{k} \binom{k}{i} = \binom{n}{i} \cdot2^{n-i}$$
Can some provide a conceptual explanation of this (not an algebraic one)?
$$\sum_{k=i}^{n} \binom{n}{k} \binom{k}{i} = \binom{n}{i} \cdot2^{n-i}$$
Can some provide a conceptual explanation of this (not an algebraic one)?
Given an $n$-element set $A$, the LHS counts the ways of picking sets $B$ and $C$ with $A\supseteq B\supseteq C$ and $|C|=i$. What if you pick $C$ first, and then count how many $B$ are there with $A\supseteq B\supseteq C$?