A problem in permutation/combination **Q.A boat is to be manned by eight men,of whom $2$ can only row on bow side and $1$ can only row on stroke side,in how many ways can the crew be arranged?** _I approached the problem in the following way:_ -Since there were $8$ crew members.$2$ of them should be assigned to bow side.$1$ applied permutation. -Now,$1$ member out of $6$ should be assigned to stroke side,I again applied permutation. No. of ways crew can be assigned to row is $8!/5!$.I think No. of ways crew can be assigned to row = No of ways can the crew be arranged Well,this does not solve the problem. Where is my mistake? [Permutation and Combination- Rowing a Boat is a similar question.But it didnot attracted any answer]

Presuming that 4 men are to be put on each side, first place 2 "special" on the bow side and 1 "special" on the stroke side. Now there are 5 left.

Choose 2 men from the 5 left for the bow side in ${5\choose2}$ = 10 ways. The remaining automatically get assigned to the stroke side.

So the answer is just 10, if order does not matter.

[If order does matter , it will be $10\cdot4!\cdot4!$ ]