A spider has 8 identical socks and 8 identical shoes. In how many different orders can she put them on, knowing that, obviously, on each leg, she must put the shoe after the sock?
Number the spider's legs from 1 to 8. Let $a_{i}$ be the action of putting a sock on the $i$-th leg, and $b_{i}$ be the action of putting a shoe on the $i$-th leg. There are 16! ways to order the 16 actions $\left(a_{i}\right)$ and $\left(b_{i}\right)$. Consider one of these orders. From this order, we allow swapping the positions of the actions $a_{i}$ and $b_{i}$ for all $i$, which allows us to reach $2^{8}$ different orders. Among these orders, only one corresponds to a correct way of putting on shoes: the order in which for all $i, a_{i}$ appears before $b_{i}$. Therefore, the spider can put on its shoes in $16!/ 2^{8}=81729648000$ different ways.