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by pkw209 » Fri Mar 12, 2010 2:44 pm
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?


a. 28
b. 32
c. 48
d. 60
e. 120

Answer is b
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by kstv » Fri Mar 12, 2010 6:08 pm
Total no of ways of arranging them without the restriction that the 2 daughters will not sit together
2 X 4! = 48. Driver in 2 ways (Father and Mother) X Arranging remaining 4 family members

Possible arrangements if the daughters sit together
Driver seat = 2 ways
Front Seat = 2 ways.(Assuming daughters are sitting together so none will sit on the front seat)
Two Daughters will sit together so they can be assumed as one entity
First arrange the Daughters as one entity and son or one of the parents = 2 but the daughter are not one entity so they can be arranged in 2 ways. Arranging 2 daughters and either the Parents or son is possible in 2 X 2
Total ways of arranging all of them when two daughters are together = 2x2x2x2= 16

So when they are not together = Total possibilities - When they are together = [spoiler]48-16 = 32[/spoiler]
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by money9111 » Fri Mar 12, 2010 7:44 pm
to this day I disagree with this answer... i'm too lazy to find exactly where i got confused but i remember thinking "but this answer doesn't count the possibilities of one of the daughters sitting in the front seat." but now that I type it out... i think i've just answered my question... because if you're trying to figure out the possibilities where they WOULD be sitting together... this would only happen in the back seat...

hahah thanks
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