To better get across what I'm trying to ask I think it would be easier to consider the following question.
Q1. A rowing boat has 8 seats with four either side, and there are 12 men to choose from. If two men only can row on one side, and three men on the other, how many possible seating arrangements are there?
My approach to this question would be to find the number of ways to choose four men for the first side from the 12 available i.e. $12$ C $4$ ways.
Now, the other for seats can be filled in $8$ C $4$ ways. Since there are four ways to arrange each chosen person on either side of the boat, there are $12$ C $4$ $\times$ $8$ C $4$ $\times (4!)^2$ ways to arrange eight men throughout the boat.
The solution provided for this question essentially 'fixes' (i.e. three men fixed on one side and two men fixed on the other) five out of the 12 men such that the number of ways to arrange the 12 men throughout the boat can be given by $7$ C $1$ $\times$ $6$ C $2$ $\times (4!)^2$.
I don't understand why we have to fix these five men. I could understand if the question said for example, "Mr. A, Mr.B...Mr.E" (five distinct men) must always row in the boat or "Mr. A and Mr. B must always row on one side" and "Mr. C, Mr. D and Mr. E must row on the other".