Question

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?A. 28B. 32C. 48D. 60E. 120

Answer 1

Answer:

B. 32

Step-by-step explanation:

Let's start by defining the multiplication principle.

If a first experiment can happen in ''a1'' ways,a second experiment can happen in ''a2'' ways ... and finally a ''ni'' experiment can happen in ''ai'' ways  ⇒ the total ways in which the whole experiment can take place are :

a1 x a2 x ... x ai

Let's also define the arrangement of ''n'' objects in a straight line.

This arrangement is the number n!

For example if we want all the arrangements for letters A,B and C (n = 3) this number is 3! = 3.2.1 = 6

For the problem we are going to calculate the arrangements in which the father is driving and then multiply that number by 2 to obtain the arrangements in which the father and the mother drive.

When the father drives there are 4 possible seats for 4 persons ⇒ 4! = 16 ways this 4 persons can sit.

Now we subtract the ways in which the two daughters are sitting next to each other.

This number is 2 x 2 x 2.

The ''first'' two is because the two daughters can seat one in the middle seat and one in the right window or one in the middle seat and one in the left window.

The ''second'' two is because the daughters are each different.

And the last two is the ways in which the mother and the son can sit ⇒

$4!-2^{3}=24-8=16$

If the father drive, there are 16 ways for this arrangement.

Now we multiply by 2 to obtain all the arrangements (we don't forget the case in which the mother drives).

$(16).2=32$

There are 32 possible seating arrangements.