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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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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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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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