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ps question

by magical cook » Sat Nov 03, 2007 8:35 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?
28
32
48
60
120
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by raulverde » Sat Nov 03, 2007 4:19 pm
I come up with 32.

Total combinations = 2( Mother or Father can drive) *4*3*2*1 = 48

Lets calculate the total number of cases when the Daughters do sit together.
2(Mother or Father can drive) * 2 (Only the other parent and Son can sit in front seat) * 4 (Treat daughters as one unit ) = 16

48 - 16 = 32
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by camitava » Sat Nov 03, 2007 11:14 pm
Agree wit hraulverde. IMO 32.
Correct me If I am wrong


Regards,

Amitava
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