Boating Probabilities..

Side A

There are $4$ seats and $3$ men must occupies them. The number of ways we can arrange them is $\frac{4!}{(4-3)!}=\binom{4}{3}3!$ (non-repeated ordered disposition)

Side B

There are $4$ seats and $2$ men must occupies them. The number of ways we can arrange them is $\frac{4!}{(4-2)!}=\binom{4}{2}2!$ (non-repeated ordered disposition)

There are $3$ men and $3$ seats left. The number of ways we can arrange them is $3!$ .

Hence, there are $N$ possible disposition for the crew, namely

$\displaystyle N=\binom{4}{3} \binom{4}{2} 2! \cdot 3!^2=\boxed{1728}$

sides for 5 women are pre-decided of the remaining 3, there are 3c1 = 3 ways to choose the woman for the bow side the remaining 2 automatically get allocated to the stroke side.

now, on each side, there can be 4! arrangements of the women, so total # of ways = 3 4! 4! = 1728 <-----------

QED