Permutation involving 4 cups and 4 saucers.

There are $3$ cups on which you can place on saucer $4$. For concreteness, let's say you put cup $3$ there. Now, you might place cup $4$ on saucer $3$, and then only cups $1$ and $2$ and saucers $1$ and $2$ remain, and there's only one way to place the cups on non-matching saucers, so in this case we have $3$ possibilities.

Alternatively, cup $4$ might be placed on either saucer $1$ or saucer $2$. Now we still have to place cups $1$ and $2$ and saucers $2$ and $3$ are available. Again, there is only one way to place the cups on non-matching saucers, so we have $2\cdot3=6$ possibilities in this case, and $9$ overall.

The generalization of this line of reasoning to $n$ cups and saucers is given here.