Vamshi8411 wrote:

In how many ways can a teacher in a kindergarten school arrange a group of 6 children (Susan, Karen, lei, Tim, joy and Zen) on 6 identical chairs in a straight line so that Susan is on the left of Tim?

A. 720

B. 360

C. 240

D. 120

E. 60

Method-1:Total ways in which 6 children can be arranged on 6 chairs = 6*5*4*3*2*1 = 6! = 720

But in half cases Susan will be left of Tim and in other half of cases Tim will be on left of Susani.e. Desired cases in which Susan is on the left of Tim = (1/2)*720 = 360

Method-2:We have six vacant chairs _ _ _ _ _ _

out of which 2 have to be occupied by Tim and Susan, No. of ways of selecting chairs for Tim and Susan = 6C2 = 15

No. of ways of arranging Tim an Susan on selected chairs = 1 (Because Susan must be on the left of Tim)

Total ways of arranging 4 remaining children on 4 remaining chairs = 4*3*2*1 = 4! = 24

i.e. Total Arrangement = 6C2 * 1 * 4! = 15*1*24 = 360

Method-3:Case 1: S _ _ _ _ _

Total ways of arranging Tim to the right of Susan = 5

Total ways of arranging Remaining 4 children = 4!

i.e. Total Arrangement in this case = 5*4!

Case 2: _ S _ _ _ _

Total ways of arranging Tim to the right of Susan = 4

Total ways of arranging Remaining 4 children = 4!

i.e. Total Arrangement in this case = 4*4!

Case 3: _ _ S _ _ _

Total ways of arranging Tim to the right of Susan = 3

Total ways of arranging Remaining 4 children = 4!

i.e. Total Arrangement in this case = 3*4!

Case 4: _ _ _ S _ _

Total ways of arranging Tim to the right of Susan = 2

Total ways of arranging Remaining 4 children = 4!

i.e. Total Arrangement in this case = 2*4!

Case 5: _ _ _ _ S _

Total ways of arranging Tim to the right of Susan = 1

Total ways of arranging Remaining 4 children = 4!

i.e. Total Arrangement in this case = 1*4!

Total Arrangements = 5*4!+5*4!+4*4!+3*4!+2*4!+1*4! = 4!(5+4+3+2+1) = 24*15 = 360

Answer: Option B

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