Four friends at five diner branches all apart

Four friends plan to have lunch at Jimmy's Diner, not realising it has five branches. Each friend independently goes to a branch chosen uniformly at random. What is the probability that no two of them are at the same branch?

$\tfrac{24}{125}$. There are $5^4=625$ equally likely assignments, of which $5\cdot4\cdot3\cdot2=120$ use four different branches. So the probability is $\tfrac{120}{625}=\tfrac{24}{125}$.