By counting in two ways, prove that $\sum_{i=0}^n \binom{i}{k} = \binom{n+1}{k+1}$ for all $k,n \ge 0$.

Question

I am thinking that the RHS can be counted as $\binom{n}{k} + \binom{n}{k+1}$. But I don't how the following counting may go.

Answer 1

The RHS represents number of such subsets $A$ of set $\{1,...,n+1\}$ that $|A| = k+1$.

We'll show, the LHS counts the same. Take any $i \in \{k,...,n\}$. We want to consider subsets $A$ (with $|A| = k+1$) of set $\{1,...,n+1\}$ for which the element $i+1$ is the biggest. So we have to choose another $k$ elements (as $i+1$ is already in our subset) from the set $\{1,...,i\}$, and assign them to the set $A$. That for a fixed $i$ gives us ${i \choose k}$ ways. Note that, every $k+1$ - subset can have any element starting from $k+1$ ending at $n+1$ as the biggest. So summing:, we get $\sum_{i=k}^n {i \choose k} = {n+1 \choose k+1}$.

And $\sum_{i=0}^n {i \choose k} = \sum_{i=k}^n {i \choose k} = {n+1 \choose k+1}$

Answer 2

Note that $$\binom {n+1}{k+1}=\binom {n}{k}+\binom {n}{k+1}$$

Also $$\binom {n}{k+1}=\binom {n-1}{k}+\binom {n-1}{k+1}$$

Apply the identities successively until you start getting zeros.

$$\binom {n+1}{k+1}=\binom {n}{k}+\binom{n}{k+1}=$$

$$\binom {n}{k}+\binom{n-1}{k}+\binom {n-1}{k+1}+...=$$

$$\binom {n}{k}+\binom{n-1}{k}+\binom {n-2}{k}+\binom {n-2}{k+1} ...=\sum_{i=0}^n \binom{i}{k}$$