A family including a father, a mother, three sons and three daughters,

Question

A family including a father, a mother, three sons and three daughters, decide to watch a movie in their home theater room. The theater room has two front seats, three seats in the second row and three seats in the third row. The right most seat in the front row is reserved for one of the parents. If all the daughters want to sit in the same row and two of their sons, William and Jacob refuse to sit next to each other, in how many ways can the family be seated in the home theater room?

A. 320
B. 380
C. 384
D. 390
E. 400

Show Answer

Official Answer

C

A. 320
B. 380
C. 384
D. 390
E. 400

GmatPoint · Accepted Answer

Since the rightmost seat of the first row is already reserved for a parent, it can be filled in two ways.
Also, as the girls have to sit together, they will occupy one whole row and hence can be seated in 2*3! ways. 3! is multiplied for the arrangement of the girls among themselves.

Now, they are left with 4 vacant seats. (1 in the first row and 3 in one of the vacant rows)

Let us consider the cases for William.
1) If he is sitting in the first row, the number of ways the other three will sit will be 3! = 6.

2)William is sitting in one of the vacant rows:
I)W_ _: For this arrangement, the middle seat of the vacant row can be filled in two ways, and the other 2 seats can also be filled in two ways. 
Thus, number of arrangements in this case = 2*2 = 4.
Similarly, for _ _ W, the number of arrangements will be 4.

II)_ W _: Jacob will sit in the front row for this arrangement. The total number of arrangements will be 2.

Thus, the total number of arrangements = 2*2*3!(3! + 4 + 4 + 2) = 24 * 16 = 384

Thus, the correct option is C.

lordwells · Answer

IMO the correct answer is "C" 384

My calculations are as follows
2 * 2 * 6 * (2 * 6 + 2 * 2)

2 ways of choosing the parent seated in the rightmost front seat (mother or father)

2 ways of choosing the row where the three daughters are seated (as a matter of fact, it could be the second or the third row)

6 ways of arranging the sisters (3!)

...times...

the sum of two different cases:

1) Either W or J are seated in the leftmost seat in the front row
Then we would have

2 ways of choosing one of them (W or J)

6 ways of arranging the remaining members of the family (that is, the one between W or J and the one between the two parents who have not previously been chosen as being seated in the front row, plus the other son) in the three-seat row we are left with

2) Neither W or J are seated in the leftmost seat in the front row and they are therefore sharing the same not-occupied-by-their-sisters three-seat row

As a consequence of the constraint which impose the two of them NOT to be seated next to each other, another family member must be seated in between
Then we would have

2 ways to choose this family member (who could essentially be the other son or the other parent who is not seated in the front row)

2 ways of arranging W and J (as a matter of fact, that row could be in the form "W-other family member-J" or "J-other family member-W")

In this way, I think all possible configurations have been exhausted and taken into account

In the end, the calculations boil down to
24 * 16 = 384

Answer "C"

Posted from my mobile device

Gemmie · Answer

(*)
3 Daughters can only sit together either in 2nd or 3rd row
In each row, there are 3! ways of arrangement
 => Total number of ways of arrangement for 3 seats (for 3 Daughters): 2*3! = 12

(**)
Either Parent must sit in the right most seat in the front row
 => 2 ways of arrangment for Right Front seat: 2

(***)
There are 4 seats (including 3 seats in a same row and 1 front seat) left for 3 Sons and Remaining Parent
=> Total ways of arrangement: 4! = 24

However, there are some arrangements where J and W sit next to each other must be removed
 
 Number the seats in the remaining 3-seat row is a1, a2, a3 
 Consider J-W sit together as one block ==> Total arrangements: 4 = 2 * 2
 
 2 ways to arrange the J-W block, either a1-a2 or a2-a3, 
 then 2 ways to arrange the remaining 1 Son and 1 Parent)   
 In a J-W block, there are 2 ways of arranging J and W 
 => Number of arrangments where J and W do not sit next to each other = 4 * 2 = 8  
 => Number of arrangments where J and W do not sit next to each other: 24 - 8 = 16

=>>>> ANSWER: 12 * 2 * 16 = 384

Aakash7 · Answer

X   P 
 X   X   X
X   X   X

Situations:
1 parent takes right most seat = 2C1 = 2
AND
3 daughters select any 1 of the 2 "3 seater" rows = 2C1 =2
AND
Arrangement of 3 daughters = 3! = 6
AND
    { Any 1 on J/W sits on 1st row = 2C1 = 2
     AND Remaining 3 People arranged in remaining 3 seats = 3! = 6  } 
  OR 
   { Any 1 of 2 people other than J/W sits in 1 row = 2C1 = 2
     AND   {  Arrangement of J/W/X  -  Arrangement such that J and W sits together  } } 
 ]

This will give:
2*2*3! * (2*3! + 2*(3!-4))
= 24 * (12 + 4)
= 24 * 16
= 384

Option C

Gimmy · Answer

I solved this by first finding the number of ways the girls can be arranged cause it is the most straightforward = 2x3! = 12. So the answer has to be divisible by 12.
Then I figured the next few steps involving the boys will be complicated so I took a glance at all the options to see if I can eliminate any options that are not divisible by 12. Just nice it leaves us with Option C.

For practice and learning purposes, this might not be the best way but just putting it out there that sometimes by analyzing and eliminating options may quickly give you the answer.

Rishm · Answer

My man, I did the same thing, I used a slight different method, where i saw there was 4 extra way to place the boys, once with one parent and once with different parent (2 parent, 2 boys) as they cant sit together, and only option which had 4 as extra was 384.

A family including a father, a mother, three sons and three daughters,

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A family including a father, a mother, three sons and three daughters,

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards