Three young brother-sister pairs from different families need to take

Question

Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?

A. 60
B. 72
C. 92
D. 96
E. 120

Show Answer

Official Answer

D

A. 60
B. 72
C. 92
D. 96
E. 120

nick1816 · Answer

There are 2 cases possible
1. When children from 3 different families sit on the second row
Number of ways possible= (2C1*2C1*2C1)*3!* 3!(1-1!+1/2!-1/3!)= 2*2*2*6*2=96
2. When children from 2 families sit in the second row
The scenario possible
Family1 Family 2 Family1
Family3 Family 2 Family 3

We can see that Whoever sits in the middle of second row, his sibling have to sit in the middle of the third row too, which contradicts our given condition.
Hence 0 case possible

Total seating arrangements are possible for this trip= 96

ChiragSabharwal · Answer

Please explain this part. How did you calculate this?

nick1816 · Answer

When children from 3 different families sit on the second row
 The number of ways are possible to select 1 child from a family= 2C1
Because you have to select 1 child from all three families, total number of ways= 2C1*2C1*2C1
Total sitting arrangements possible in the second row= (2C1*2C1*2C1)3!

Now 3 children left and 3 seats left in the third row but we have a constraint that no child may sit directly in front of his or her sibling.
It's similar to de-arrangement problem, when no elements of a set left in it's original place. The formula for the de-arrangements for a set of n elements is

Number of de-arrangements possible= n!(1-1!+1/2!-1/3!......)

In our case there are 3 elements
Hence total de-arrangements possible= 3!(1-1!+1/2!-1/3!)=2

ChiragSabharwal · Answer

Thanks for your explanation, but is de-arrangement really tested in GMAT? Can't we just say that there are two rows, so we will multiply the selected and arranged possible figures of people in the second row by 2 and get the answer?

nick1816 · Answer

You actually don't have to use de-arrangement formula, but if you are familiar with it, then why not?
There are only arrangements possible in the back row

x y z 
z x y

and

x y z
y z x

You can use either approach!!!

Thanks for your explanation, but is de-arrangement really tested in GMAT?

rocksailor9 · Answer

Scenario 2 is OK.

But in scenario 1, why are we not taking row selection as well.

1. When children from 3 different families sit on the second row
Number of ways possible= (2C1*2C1*2C1)*3!* 2 (arrangements possible in the other row)*  2 (FOR EITHER CHOOSING row 2 or 3) 
= 192. 
Is the OA correct?

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Three young brother-sister pairs from different families need to take

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