How do I do questions which ask for distinct people to sit on a number of seats? A minibus has 17 seats arranged in 4 rows. The back row has 5 seats and the other 3 rows have 2 seats on each side. 11 passengers get on the minibus.
(i) How many possible seating arrangements are there for the 11 passengers? 
(ii) How many possible seating arrangements are there if 5 particular people
sit in the back row?
Of the 11 passengers, 5 are unmarried and the other 6 consist of 3 married couples.
(iii) In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples and 1 other person, who may or may not be married?

I am stuck in part (ii).
What I have done so far:
Step 1: Determine how many ways to get 5 distinct people to sit in the back seat. so it is 11P5 = 55440
Step 2: Determine ways to seat the remaining 6 people (11-5 = 6 people) at remaining seats (17 - 5 = 12 seats) so 12P6 = 665280
I used P because the ordering of the people matter. 
But if I multiply 1 and 2, it seems like a ridiculously large number. 

I am stuck in part (iii).
What I have done:


*

*Choose 2 couples from 3 couples.
3P2 = 6

*Choose the last one person from remaining people.
remaining people = 1 couple + 5 singles = 2 + 5 = 7
So 7P1 = 7

*Multiply:
6*7=42 
But the answer is 21.  Why is it using C instead of P? Shouldn't the order matter. Eg: couple 1 and couple 2 is the same as couple 2 and couple 1. 
 A: 
A minibus has $17$ seats arranged in $4$ rows. The back row has $5$ seats and the other $3$ rows have $2$ seats on each side. $11$ passengers get on the minibus. How many possible seating arrangements are there if $5$ particular people sit in the back row?

If we are told five particular people sit in the back seat, we do not need to select them.  In effect, we are being told that Abigail, Bruce, Claire, David, and Esme sit in the back seat. We just have to arrange them in the back seat, which can be done in $5!$ ways.  The remaining six people can be seated in the remaining twelve seats in $P(12, 6)$ ways.  Hence, the number of seating arrangements of $11$ passengers in the $17$-seat minibus if five particular people sit in the back seat is $5!P(12,6)$.

Of the $11$ passengers, $5$ are unmarried, and the other $6$ consist of $3$ married couples.  In how many ways can $5$ of the $11$ passengers on the bus be chosen if there must be $2$ married couples and $1$ other person, who may or may not be married?

We must select two of the three couples, which can be done in $\binom{3}{2}$ ways, and one of the remaining seven people, which can be done in $\binom{7}{1}$ ways.  Hence, there are $$\binom{3}{2}\binom{7}{1}$$ possible selections of two married couples and one other person, who may or may not be married.
The order in which the couples are selected does not matter.  If we first choose the couple Abigail and Bruce and then choose the couple Claire and David, we will select the same four people as if we first choose the couple Claire and David and then choose the couple Abigail and Bruce.
