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Probabilty Question

Expert replies
by Kaunteya » Sun Mar 02, 2008 3:05 pm
The question goes something like this:

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

Kaunteya
Last edited by Kaunteya on Sun Mar 02, 2008 5:22 pm, edited 1 time in total.
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Source: — Problem Solving |

by gmatguy16 » Sun Mar 02, 2008 5:15 pm
imo 32.,i would just just list down the possibilities.assume dad(d) is driving
d m
g1 s g2


d s
g1 m g2


d g1
s m g2

now for first instance we have 2 different possibilities since we can switch g1 and g2 ,for second instance we have 2 again,for third case we have 6 combinations with g1 in front seat and 6 with g2 in front seat...
total = 16,16 repeat for mom driving
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