$\binom{n}{k} = \binom{n}{n-k}$. This identity reflects

1. SYMMETRY of binomial coefficients: choosing $k$ from $n$ to INCLUDE is equivalent to choosing $n-k$ to EXCLUDE. 2. So both sides count the same thing — they must be equal. 3. Algebraic verification: $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ and $\binom{n}{n-k} = \dfrac{n!}{(n-k)!\,k!}$. Identical. 4. Pascal's identity (option A) is the different statement $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$. 5. Options C, D refer to other counting principles. _Source: Oscar Levin, "Discrete Mathematics: An Open Introduction", §1.2 (Binomial Coefficients — Pascal's Triangle and symmetry)._