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## Difficult Prob Question

##### This topic has expert replies
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### Difficult Prob Question

by chaitanya.mehrotra » Mon Jul 25, 2011 10:30 am
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?
28
32
48
60
120

OA 32

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by winniethepooh » Mon Jul 25, 2011 12:18 pm
You can consider simply counting with any parent driving and then adjusting the other 4 family members in 16 ways then multiply the resultant by 2 (which is for the second parent driving).

Or you can consider all ways without any restrictions then subtracting the possibilities with the sisters sitting only together.
Frong driver seat = 2 ways
Front passenger seat= 4 ways
Three back passenger seats in 3 * 2 * 1 ways.

Total = 2*4*3*2*1 = 48 ways._____(1)

Two sisters can sit together only in the back seat:
The arrsngements will now be:
Front driver seat = 2 ways.
Front passenger seat = 2 ways
Back seat can be filled by considering both daughters as 1 unit = 2c1 ways = 2 ways
As both daughters can interchange their seats = 2ways.

Total = 2*2*2*2 = 16 ways.___(2)

Number of ways with restrictions = 48-16 = 32 ways.____(1)-(2)

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by ECMoyano » Mon Jul 25, 2011 12:39 pm
You can consider simply counting with any parent driving and then adjusting the other 4 family members in 16 ways then multiply the resultant by 2 (which is for the second parent driving).

Or you can consider all ways without any restrictions then subtracting the possibilities with the sisters sitting only together.
Frong driver seat = 2 ways
Front passenger seat= 4 ways
Three back passenger seats in 3 * 2 * 1 ways.

Total = 2*4*3*2*1 = 48 ways._____(1)

Two sisters can sit together only in the back seat:
The arrsngements will now be:
Front driver seat = 2 ways.
Front passenger seat = 2 ways
Back seat can be filled by considering both daughters as 1 unit = 2c1 ways = 2 ways
As both daughters can interchange their seats = 2ways.

Total = 2*2*2*2 = 16 ways.___(2)

Number of ways with restrictions = 48-16 = 32 ways.____(1)-(2)
Interesting, I did something similiar

Who can drive: 2 (mom or dad)
Who can ride shotgun: 4 (whoever isn't driving)

Than I labeled backseats A B C and gave A and C the value of the sisters.
AB, AC, BA, BC, CA, CB, and than eliminated AC + CA since they can't sit next to each other leaving me with 4 possible combinations.

2*4*4 = 32

Just wondering, how did you get Front driver and Front Passenger to be both 2?

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by winniethepooh » Mon Jul 25, 2011 1:32 pm
Front Driver = Mom or dad
Front Passenger: = Son or Father(This is to satisfy the condition that both sisters are sitting together, and theycan sit together only at the back seats).

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by GMATGuruNY » Mon Jul 25, 2011 8:37 pm
chaitanya.mehrotra 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?
28
32
48
60
120

OA 32
An alternate approach:

One daughter in the front passenger 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 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.

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by winniethepooh » Mon Jul 25, 2011 8:56 pm
Mitch can we solve it by the normal way, that is considering 4 options in the front passenger seat and then considering the restrictions at the back seat.
I understand what you've done is similar, but can we conclude it without having to add up two individual answers?

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by jainpiyushjain » Sun Mar 31, 2013 4:31 am
GMATGuruNY wrote:
An alternate approach:

One daughter in the front passenger 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 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.