seating arrangements

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gmatnmein2010: Master | Next Rank: 500 Posts; Posts: 185; Joined: Thu Feb 04, 2010 3:06 am; Thanked: 6 times

by gmatnmein2010 » Mon Feb 08, 2010 5:33 am

1. 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

i took a lot of time and cud not solve it
pls suggest a short method
thanks

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Source: Beat The GMAT — Problem Solving | Mon Feb 08, 2010 5:33 am

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Mon Feb 08, 2010 6:02 am

gmatnmein2010 wrote:1. 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

i took a lot of time and cud not solve it
pls suggest a short method
thanks

Case 1:-Only 1 parent driving+ 1 daughter in front seat . 2C1 x 2C1(selection of front seat daughter) x 3!(back seat arrangement)=24
Case 2 :-Both parents in the front seat. 2C1(for driver) x 2!(daughters arrangementson always sits in middle)=4
Ans. is 28.A

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thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Mon Feb 08, 2010 6:08 am

Total ways to seat them with one parent driving: 2 options for driver * 4 options for other seats = 2*4*3*2*1 = 48.
Assuming the 2 sisters do sit together, front seat we have 2 options for driver and 2 options for passenger (the other parent + son) = 4 ways.
Back seats: sisters sit together so either they sit in seats 1 and 2 or seats 2 and 3 = 2 ways. Also sisters themselves can sit in 2 ways so total = 2*2 = 4. So total ways sisters can sit together = 4*4 =16.
So ways they wont sit together = 48-16 = 32 ways.

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Mon Feb 08, 2010 6:12 am

gmatnmein2010 wrote:1. 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

i took a lot of time and cud not solve it
pls suggest a short method
thanks

there can be two cases
Case 1: Only 1 parent driving + 1 daughter in front seat .
No. of combination= 2C1 x 2C1 x 3!=24
Case 2 : Both parents in the front seat. and in back seat girls will at extreme and son in between = 2C1 x 2!=4
Ans. is 28A

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