Seating question

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vaibhavjha: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Thu Aug 19, 2010 7:26 am; Thanked: 1 times

by vaibhavjha » Fri Nov 12, 2010 8:16 am

In how many ways can a teachers seat 6 children (Susan,tim,john,greg,Paul,Paris) in 6 chars in a straight line so that Susan is always on the left of Tim ?

A. 60
B. 120
C. 360
D. 720
E. 1200

OA: C

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Source: Beat The GMAT — Problem Solving | Fri Nov 12, 2010 8:16 am

beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Fri Nov 12, 2010 8:44 am

vaibhavjha wrote:In how many ways can a teachers seat 6 children (Susan,tim,john,greg,Paul,Paris) in 6 chars in a straight line so that Susan is always on the left of Tim ?

A. 60
B. 120
C. 360
D. 720
E. 1200

OA: C

Fix susan at 1st position , tim can sit to left in 5 ways.
Fix susan at 2nd position , tim can sit to left in 4 ways.
Fix susan at 3rd position , tim can sit to left in 3 ways.
Fix susan at 4th position , tim can sit to left in 2 ways.
Fix susan at 5th position , tim can sit to left in 1 ways.

Add = 5+4+3+2+1 = 15.
remaining 4 can be arranged in 4! ways.
Total ways - 15*4! = 360.

Please use spoilers for OA.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Nov 12, 2010 9:06 am

vaibhavjha wrote:In how many ways can a teachers seat 6 children (Susan,tim,john,greg,Paul,Paris) in 6 chars in a straight line so that Susan is always on the left of Tim ?

A. 60
B. 120
C. 360
D. 720
E. 1200

OA: C

Total number of ways to arrange 6 elements is 6! = 720.

The probability that Susan sits to the left of Tim is the same as the probability that she sits to the right of Tim. Thus, in half of the arrangements, Susan will sit to the left of Tim; in the remaining half, she will sit to the right of Tim.

1/2 * 720 = 360.

The correct answer is C.

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