Basic combinatorics proofs

\(\displaystyle \binom{n + 1}{2} - \binom{n}{2} = n\)
\(\displaystyle \binom{n}{m}\binom{m}{k} = \binom{n}{k}\binom{n - k}{m - k}\)
For \(0 \le k \le n\),  \(\displaystyle n! = \binom{n}{k}k!(n - k)!\).
For \(0 \le k \le n\),  \(\displaystyle \binom{n}{k} = \binom{n}{n - k}\).
For \(k, n \ge 0\),  \(\displaystyle \left({\mkern-5mu}{\binom{n}{k}}\mkern-5mu\right) = \binom{n + k - 1}{k}\).
For \(0 \le k \le n\),  \(\displaystyle \sum_{k = 0}^n \binom{n}{k} = 2^n\).
For any \(x, y \in \mathbb{N}\) and \(n \ge 1\),  \(\displaystyle (x + y)^n = \sum_{k = 0}^n \binom{n}{k}x^ky^{n - k}\).
For \(0 \le k \le n\),  \(\displaystyle \sum_{k = 0}^n \binom{n}{2k} = 2^{n - 1}\).
For \(0 \le k \le n\),  \(\displaystyle \sum_{k = 0}^n \binom{n}{2k} = 2^{n - 1}\).
For \(0 \le k \le n\),  \(\displaystyle k\binom{n}{k} = n\binom{n - 1}{k - 1}\).
For \(0 \le m \le k \le n\),  \(\displaystyle \binom{n}{k}\binom{k}{m} = \binom{n}{m}\binom{n - m}{k - m}\).
\(\displaystyle \binom{2n}{2} = 2\binom{n}{2} + n^2\)
Algebraic proof