Pascal's identity

Step 1 (fix one element and split by membership): fix one object $x$ among the $n$ objects. Every $k$-element subset either contains $x$ or does not; these two cases are disjoint and together cover all $C_n^k$ subsets.

Step 2 (count subsets containing x): a subset containing $x$ is formed by choosing the remaining $k-1$ elements from the other $n-1$ objects, giving $C_{n-1}^{k-1}$ such subsets.

Step 3 (count subsets not containing x): a subset avoiding $x$ is formed by choosing all $k$ elements from the remaining $n-1$ objects, giving $C_{n-1}^k$ such subsets.

Step 4 (apply the addition rule): since the two cases are disjoint and exhaust all size-k subsets, the addition rule gives $C_n^k = C_{n-1}^{k-1} + C_{n-1}^k$.