This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 156
Joined: 15 Jul 2012
Location: New York, USA
Thanked: 34 times
Followed by:1 members

Sedan Holiday

by kartikshah » Mon Jul 23, 2012 7:36 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

OA is 32.

I got 48 ways to seat people (2*4*3*2*1) ways. However I could count only 12 seating arrangements to deduct from 48 for those events in which daughters occupied adjacent seats.

Could someone clarify?

User avatar
Community Manager
Posts: 991
Joined: 23 Sep 2010
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Mon Jul 23, 2012 7:58 am
Total Ways = Driver seat (2 ways) * 4 * 3 * 2 = 48 ways
Say the two sisters sit together S1 and S2
S1,S2 sequence possible ways = Driver seat (2 ways) * Next Front Seat (2 ways other parent and son) * Back remaining 1seat (2 ways son and other parent) * rest 2 back seats (1 way) = 8 ways
Now, change the arrangement of Sisters to S2,S1 = 8 ways
Total number of ways when both sisters sit together = 2*8 = 16 ways

Number ways they do not sit together = 48 - 16 = 32 ways
If the problem is Easy Respect it, if the problem is tough Attack it

Master | Next Rank: 500 Posts
Posts: 156
Joined: 15 Jul 2012
Location: New York, USA
Thanked: 34 times
Followed by:1 members

by kartikshah » Mon Jul 23, 2012 8:20 am
I see the mistake in my working now.
I had not calculated 4 possible ways in which sisters would sit next to each other in the front seats.
So 12+ 4 = 16

!!
Thanks