Problem Solving

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
[spoiler]
Ans: B[/spoiler]

Can this be done using circular arrangements (n-1)!/2 with a twist? The MGMAT explanation takes 15 minutes to read - let alone to really be able to do on the test...

Are you sure 32 is the OA? Pls check the answer once again.

benjiboo 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
[spoiler]
Ans: B[/spoiler]

Can this be done using circular arrangements (n-1)!/2 with a twist? The MGMAT explanation takes 15 minutes to read - let alone to really be able to do on the test...

b]One daughter in the front passenger seat:[/b]
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the front passenger seat, number of choices = 2.
Number of ways to arrange the 3 remaining people = 3! = 6.
Multiplying the results above, we get:
Number of arrangements = 2*2*6 = 24.

The two daughters separated in the back seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the leftmost back seat, number of choices = 2.
Since the 1 remaining daughter must be in the rightmost back seat, number of choices = 1.
Number of ways to arrange the 2 remaining people = 2! = 2.
Multiplying the results above, we get:
Number of arrangements = 2*2*1*2 = 8.

Thus, total arrangements = 24+8 = 32.

The correct answer is B.

driver --> mom or dad, 2! ways
2nd front seat --> parent 1 way // three back seats daugter-son-daughter where for daughters 2! ways // total 2!*2!=4 ways

driver --> mom or dad, 2! ways
2nd front seat --> son 1 way // three back seats daugter-parent-daughter where for daughters 2! ways // total 2!*2!=4 ways

driver --> mom or dad, 2! ways
2nd front seat --> daughter 2! ways // three back seats one daughter, one parent, one son or 3! ways // total 2!*2!*3!=24 ways

grand total 4+4+24=32 ways

benjiboo 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

Right... But they are sort of sitting in a circle if you think about it... so I am wondering if there was a way to use that concept to figure this out. Circular arrangements.

benjiboo wrote:Right... But they are sort of sitting in a circle if you think about it... so I am wondering if there was a way to use that concept to figure this out. Circular arrangements.

We cannot look at this as circular permutation with the restrictions given. The very essence of circular permutation is similarity of orders arranged in the clock-wise and the counter clock-wise directions. Unfortunately if you pick a person sitting in the middle of the back seats that person may not be the first person to start our arrangement and be considered equivalent to the order with the person (parent here) sitting in the front seat always.

Another method can also be implemented. It goes as below..

At Driver Seat - Option are two so no. of ways are 2!(Either Father or Mother)
Rest all possibilities are - 4!
So Total no. of Possibilities = 4! * 2! = 48

Now,

Group the two daughters together as one.

So at driver seat - Option are two so no. of ways are 2!(Either Father or Mother)

At Second front seat - Only Two Options i.e Either father/mother or Son because the two daughters are combined as one so cant seat at second front seat - so no. of ways are 2!

At back seat 3 persons are to be seated but group is only two since the duo daughters are combined as one - so no. of ways are 2!

and lastly the two daughters can be arranged in 2! ways.

so total ways in which daughters are together are 2! * 2! * 2! * 2! = 16

So the no. of cases in which daughters are not together are 48 - 16 = 32.

Hope this may suffice the requirement.