Arrange $n$ people in a row with conditions Let $n$ people, including Alice, Bob and Eve. How many ways are there to arrange those $n$ people in a row, such that Alice is left to Eve and Bob is right to Eve. My (wrong) thought: We choose places for Alice, Bob and Eve: $C(n, 3)$. There is only one valid arrangement. Then, we multiply by $(n-3)!$ for the rest. The right answer is: $$\left( {\matrix{ n \cr 2 \cr } } \right) \cdot \left( {\matrix{ {n - 2} \cr 2 \cr } } \right) \cdot (n - 4)! \cdot 1 \cdot 1$$ Why?

Fisrt: I'm pretty sure that your answer is correct.

Second: the "correct" answer makes no sense, since the problem doesn't tell that there are actually more than three people in the row. Try to substitue $n=3$ and you get the "binomial coefficient" $\binom 12$.