A boat is to be manned by 8 men,of whom 2 can only row on bow side & 1 can only row on stroke side;in how many ways can the crew be arranged?

I tried it by selecting 2 men out of 8 for bow side,and then arrange them in 2! ways.This can be done in$\binom{8}{2}$*2! ways,and the stroke side can be crewed in 6 ways.So the required no. of ways are $\binom{8}{2}$*2!*6=336. But,my answer is not matching.It's correct answer is 5760.

See we don't want to select people we want to arrange people. Hence total ways are $^4P_2\times$$^4P_1\times$$^5P_5=5760$ assuming boat has 4 rowing on each side.
You can select which two of the five who can row either side should row Bow in ${5\choose2}=10$ ways. That determines which four row Bow an which four row Stroke. Now you have $4!$ ways of arranging the Bow side and $4!$ ways of arranging the Stroke side. So total $10\cdot4!\cdot4!=5760$.
Alternately, select seats for the three primadonnas first, which can be done in $4\cdot3\cdot 4$ ways. Afterwards the 5 remaining rowers can be seated in $5!$ ways, for a result of $$4\cdot 3\cdot 4\cdot 5!=5760$$