pratikbais wrote:
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. 28
B. 32
C. 48
D. 60
E. 120
Official Solution (Credit: Manhattan Prep)
The easiest way to solve this question is to consider the restrictions separately. Let’s start by considering the restriction that one of the parents must drive, temporarily ignoring the restriction that the two sisters won't sit next to each other.
This means that…
2 people (mother or father) could sit in the driver’s seat
4 people (remaining parent or one of the children) could sit in the front passenger seat
3 people could sit in the first back seat
2 people could sit in the second back seat
1 person could sit in the remaining back seat
The total number of possible seating arrangements would be the product of these various possibilities: 2 × 4 × 3 × 2 × 1 = 48
We must subtract from these 48 possible seating arrangements the number of seating arrangements in which the daughters are sitting together. The only way for the daughters to sit next to each other is if they are both sitting in the back.
This means that…
2 people (mother or father) could sit in the driver’s seat
2 people (remaining parent or son) could sit in the front passenger seat
Now for the back three seats we will do something a little different. The back three seats must contain the two daughters and the remaining person (son or parent). To find out the number of arrangements in which the daughters are sitting adjacent, let’s consider the two daughters as one unit. The remaining person (son or parent) is the other unit. Now, instead of three seats to fill, we only have two "seats," or units, to fill.
There are 2 × 1 = 2 ways to seat these two units.
However, the daughter-daughter unit could be d1d2 or d2d1
We must consider both of these possibilities so we multiply the 2 by 2! for a total of 4 seating possibilities in the back.
We could also have manually counted these possibilities:
d1d2X, d2d1X, Xd1d2, Xd2d1
Now we must multiply these 4 back seat scenarios by the front seat scenarios we calculated earlier:
(2 × 2) × 4 = 16
front back
If we subtract these 16 "daughters-sitting-adjacent" scenarios from the total number of "parent-driving" scenarios, we get: 48 – 16 = 32
The correct answer is B.