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by ruplun » Mon Aug 01, 2011 11:34 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?
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by winniethepooh » Mon Aug 01, 2011 12:19 pm
We can calculate this by finding the number of ways the family can sit without any restrictions and then deducting the number of ways the daughters can sit together(only)

Clculating the number of ways w/0 restrictions:
2 ways for front driver seat
4 ways for the front passenger seat
3 ways for left back seat
2 ways for mibble back seat and
1 way for the left back seat
= 2*4*3*2*1 = 48 ways.

Calculating the number of ways the daughter can sit together.
They can sit together only in the back seat:Hence, the daughters will only be placed at the back.
2 ways for front driver seat
2 ways for front passenger seat(the son or one of the parents)
2 ways for the back seats.(2 daughter sit together so they can be considered 1 unit and the remaining of son or any parent)
1 way for the remaining(Either 2 daughter or son or any parent.
total ways = 2*2*2*1 = 8 ways.
but you have to multiply this number by 2 (the number of ways the daughters can arrange amongst themselves.
So, total number of ways = 2*8 = 16 ways.
The number of ways with restrictions = 48-16 = 32ways.
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by GMATGuruNY » Mon Aug 01, 2011 2:24 pm
I posted a solution here:

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