GMATinsight wrote:

6 horses {A,B,C,D,E,F} Participate in a race. If there are no draws in the rate then in how many ways can the race end such that horse A always finishes ahead of C and B both the horses?

A) 720

B) 360

C) 240

D) 120

E) 48

Source:

http://www.GMATinsight.com METHOD 1:We need 3 places for A B and C which can be selected in 6C3 ways

Now A has to seated on the first selected place and B and C can exchange position on the remaining two positions in 2! ways

Remaining three individuals can occupy the positions in 3! ways

Total ways of arranging them as desired = 6C3 * 2! * 3! = 20*2*6 = 240

METHOD 2:A has equal chances of being ahead of B and C as B and C have to be ahead of other two hence

probability of A to be ahead of both B and C = 1/3

Total ways of arranging the six individuals = 6! = 720 ways

Favourable cases = (1/3)*720 = 240

METHOD 3:If A comes at first place A - - - - - then arrangements = 5!

If A comes at Second place - A - - - - then arrangements = 4C2*2!*3! = 72

If A comes at Third place - - A - - - then arrangements = 3C2*2!*3! = 36

If A comes at Forth place - - - A - - then arrangements = 2C2*2!*3! = 12

Total ways = 120+72+36+12 = 240

Answer: Option C

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