And 3!(8!/3!) is simply 8!, which is the total number of ways you can arrange the crew without taking the constraints into account.
Let's try to get some intuition first... let B denote someone who can only row on the bow side, S on the stroke side, and A someone who can be on either side.
One possible arrangement is:
A A
B A
B S
B S
You need to realize there are two possible ways of permuting this
(1) you can move the A's around on the left hand side or the right hand side, e.g.
B A
A S
B A
B S
So you will have to figure out the number of permutations this gives you.
(2) in addition, all the rowers are individuals whose order matters. So for any variation you get from the above, you can move the B's, the S's and the A's, e.g. if we number the ladies then the first layout above can either correspond to
A1 A2
B1 A3
B2 S1
B3 S2
or to
A2 A3
B2 A1
B3 S2
B1 S1
or any other combination which again leads to (A, B, B, B; A, A, S, S) when you remove the numbers.