Problem - 2524
Seven people form a line. If $A$ must stand next to $B$, and $C$ must stand next to $D$, how many possibilities are there?
We can employ the bundling technique to treat $A$ and $B$ as one bundle, $C$ and $D$ as another bundle. Now, we have five "super" person to form a line. There are $5!$ ways to do this. Meanwhile, $A$ and $B$ can switch position with their bundle, so do $C$ and $D$. Therefore, the final answer is $$5!\times 2! \times 2!=\boxed{480}$$