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Re: A family including a father, a mother, three sons and three daughters, [#permalink]
lordwells wrote:
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


In deed the answer is C as per the book. Thank you!
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Re: A family including a father, a mother, three sons and three daughters, [#permalink]
(*)
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
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A family including a father, a mother, three sons and three daughters, [#permalink]
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
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A family including a father, a mother, three sons and three daughters, [#permalink]
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