gmatophobia wrote:
PS Question 1 - June 5 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 Source: Manhattan | Difficulty: Hard
Great question!
Here is how I reasoned through it:
Driver Seat:
You can only have two options - either Mother or Father.
So, 2C1=2
Now we have to consider two diverigent cases:
gmatophobia wrote:
PS Question 1 - June 5 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 Source: Manhattan | Difficulty: Hard
Great question!
Here is how I reasoned through it:
Driver Seat:
You can only have two options - either Mother or Father.
So, 2C1=2
Now we have to consider two divergent cases:
Case 1: One of the two daughter sits in the front seat
If this is the case, there are
2c1 ways for this to occur (2)
And because one of the daughters is already in the front, there no restrictions on the back
So, 3! for the back.
Case one gives us: 2*2*6=24 total cases.
Case 2:
Either the Son or the other parent sits on the front seat
This is gain, 2C1 ways to do it (so 2)
However in the back, we must consider the restrictions.
The total number of ways to do it is 3! (6) like before,
However we must consider the cases where the daughters end up together (2!*2 cases) leaving us with 2 acceptable options in the back seat.
So case 2 gives us a total of 2*2*2, or 8 ways to seat.
24+8=32, which is the total acceptable number of seating arrangements