Pascal's rule

Computation.
$$\binom{n-1}k+\binom{n-1}{k-1}=\frac{(n-1)!}{k!\,(n-1-k)!}+\frac{(n-1)!}{(k-1)!\,(n-k)!}=\frac{(n-1)!}{k!\,(n-k)!}\big[(n-k)+k\big]=\frac{n!}{k!\,(n-k)!}.$$

Counting. Let $E$ have $n$ elements and fix $x\in E$. The $k$-element subsets of $E$ split into two groups: those not containing $x$, which are $k$-element subsets of $E\setminus\{x\}$, so $\binom{n-1}k$ of them; and those containing $x$, obtained by adding $x$ to a $(k-1)$-element subset of $E\setminus\{x\}$, so $\binom{n-1}{k-1}$ of them. Together there are $\binom nk$.