Problem: Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people such that exactly one person receives the type of meal ordered by that person. Solution: Consider this scenario as a string initially $BBBCCCFFF$ where the first three indices of the string denote the spots that ordered beef, the next three denote the people who ordered chicken, and the final three denote those who ordered fish. A string that we want to count may then look, for instance, like $BCFBBFCBC$ because only the first $B$ is a meal properly given according to what that index ordered. We may try to be clever and count this using binomial coefficients, but we will be disappointed quickly, as once we assign a misplaced meal, placing any other misplaced meal depends on where the others were assigned. If we just played with a few test cases, we would see that casework is very viable here. Without loss of generality, assume that one of the people who ordered beef got their order (meaning we need to place the 2 beefs among the people who ordered chicken and fish, we must place the 3 chickens among the remaining people who ordered beef and those who ordered fish, and we must place the 3 fish meals among the remaining people who ordered beef and those who ordered chicken). We can case on the remaining people who ordered beef. * Case 1: The beef orders are $BCC$ in some order. Then, all fish orders are forced to go to the people who ordered chicken and the people who ordered fish get $BBC$ in some order. There are three ways to order the beef and fish ordering individuals, so this case contributes $9$. * Case 2: The beef orders are $BFF$ in some order. This case is symmetric to the previous and hence contributes $9$. * Case $3$: The beef orders are $BCF$ in some order. We then find that the chicken orders are $FFB$ in some order and the fish orders are $CC B$ in some order. This case contributes a total of $3! \cdot 3 \cdot 3 = 54$. In total, when the beef gets a correct order, there are $9+9+54 = 72$ such orderings. Now, we multiply this by $3$ to get the final result of $216$.