# Probability that each male sits opposite to a female in two opposite 4-seat rows

$$4$$ males and $$4$$ females are randomly directed to $$8$$ seats, arranged into $$2$$ opposite rows (each row has $$4$$ seats). What is the probability that each male sits opposite to a female?

The key is $$\dfrac{16}{35},$$ but mine is only half of that. I'm not sure what I got wrong.

There are $$8!$$ ways to arrange the group. Let $$A$$ and $$B$$ be the two rows. There are $$8$$ ways to select a person for the first seat in row $$A$$ and $$4$$ ways to select a person from the other sex for the opposite seat in row $$B$$. Similarly, we have the next pairs $$(6; 3), (4; 2), (2; 1)$$ to fill up the rest of the seats, so the probability is: $$P = \dfrac{8.4.6.3.4.2.2.1}{8!} = \dfrac{8}{35}.$$

What am I missing ?

After you fill the first seat, the probability that you're still alive after filling the seat across from it is $$\frac 47$$ because of the $$7$$ remaining people, $$4$$ are of the opposite sex from the first seat. Similarly, assuming you're still alive, the probability of staying alive after filling the next pair of seats is $$\frac 35$$ and the conditional probability that you're still alive after filling the third pair of seats is $$\frac 23$$.
Multiplying these three probabilities gets you $$\frac{8}{35}$$.