Question

A car will hold 2 persons in the front seat and 1 in the rear seat. If among 6 persons only two can drive, number of ways in which car can be filled is
(a) 10
(b) 18
(c) 20
(d) 40

Answer 1

Start by finding the way of filling the driver seat first. Once you have filled the driver seat, you are left with a front seat and a back seat. It is given that only 2 of the 6 persons can drive, so there are 2 options for filling the driver seat. So, after filling the driver seat, we are left out with 5 persons and 2 different types of seats, so we will select 2 out of the 5 persons and multiply the result by 2! As the selected two will have 2 factorial arrangements in the 2 types of seats.

Complete step-by-step answer:
Let us start the solution to the above question by finding the way of filling the driver seat first. It is given that only 2 of the 6 persons can drive, so there are 2 options for filling the driver seat. Once we have filled the driving seat, we are left with the front seat and the backseat and also with 5 persons
to fill the seats. So, we will select 2 out of the 5 persons, i.e., $^{5}{{C}_{2}}$ and multiply this by 2! As the selected persons have 2! Arrangements in the two different seats. So, the final result is the multiplication of the ways of filling the driving seat and the ways of filling the other two seats.
$\text{possible arrangements}=2{{\times }^{5}}{{C}_{2}}\times 2!=2\times \dfrac{5!}{2!3!}\times 2!=40$

So, the correct answer is “Option (d)”.

Note: It is an important thing to fill the driving seat first as all the constraints are related to the driving seat. If you fill the other seats first you will have to make several cases, for example one of the people sitting among the 2 seats is a person who knows to drive and other cases as well, which would complicate the question. Also, remember that the front seat and back sit are different, so 2! Need to be multiplied to compensate for the arrangement as we did in the above question.