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.

Please help me in pointing out my mistake.
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 $10\cdot4!\cdot4!=5760$.
My answer is best as it is using basic principle
Consider 4 each on both sides
Then for 1st side we require 2 people so no. Of ways is 4(for 1st person)*3(for other one)
Then for 2nd side we require 1 person so he has 4 choice
Then for remaining 5 places we have 5*4*3*2*1 this (arrangement is of remaining people)