Example: $m$ white balls are arranged in a row, and $n$ balls are chosen from them to be painted black. If no two black balls can be adjacent, how many different ways are there to paint them?
If the balls are already painted, imagine that each of the $n$ black balls, except the last one, "eats" the white ball immediately following it. The result is $m-(n-1)=m-n+1$ remaining balls, of which $n$ are black. This produces a mapping $f$ from the set $A=\{$ ways to select $n$ black balls from a row of $m$ white balls such that no two black balls are adjacent $\}$ to the set $B=\{$ ways to select $n$ black balls from a row of $m-n+1$ white balls $\}$. This mapping is clearly injective. Conversely, for $y \in B$, let each of the $n$ black balls in $B$, except the last one, "spit out" a white ball as its right neighbor, which produces an element $x \in A$. Clearly, the image of $x$ is $y$. Therefore, $f$ is also surjective.
Hence, $f$ is a bijection, so $|A|=|B|=C_{m-n}^{m}$.