Six booths, eight visitors, none left idle

Question

A fair has 6 booths. Eight visitors each independently walk up to a booth chosen uniformly at random. What is the probability that every one of the 6 booths gets at least one visitor?

Accepted Answer

There are 6^8 = 1679616 equally likely assignments of visitors to booths. Count those that leave no booth empty using inclusion-exclusion over the booths that might be missed: banning a set of k booths forces everyone into the remaining 6-k, giving (6-k)^8 assignments, and \binom{6}{k} ways to choose the banned booths. The count hitting all booths is \sum{k=0}^{6} (-1)^k \binom{6}{k}(6-k)^8. Working the terms: 6^8 - \binom{6}{1}5^8 + \binom{6}{2}4^8 - \binom{6}{3}3^8 + \binom{6}{4}2^8 - \binom{6}{5}1^8, which is 1679616 - 2343750 + 983040 - 131220 + 3840 - 6 = 191520. Therefore P(	ext{all bo…

Six booths, eight visitors, none left idle

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