In a race of 6 horses A, B, C, D, E and F, what is the probability of

METHOD-1:

If A comes at 1st position i.e. Positions are A - - - - - then B can take position in 5 ways and C can take position in 4 ways and rest in 3! ways = 5*4*3!

If A comes at 2nd position i.e. Positions are - A - - - - then B can take position in 4 ways and C can take position in 3 ways and rest in 3! ways = 4*3*3!

If A comes at 3rd position i.e. Positions are - - A - - - then B can take position in 3 ways and C can take position in 2 ways and rest in 3! ways = 3*2*3!

If A comes at 4th position i.e. Positions are - - - A - - then B can take position in 2 ways and C can take position in 1 ways and rest in 3! ways = 2*1*3!

Total Favourable ways of arrangements = 5*4*3! +4*3*3! +3*2*3! + 2*1*3!= 3! (20+12+6+2) = 6*40 = 240
Total Arrangements of 6 horses = 6! = 720

Probability = 240/720 =1/3

METHOD-2:

Select 3 places for A B and C out of 6 positions in 6C3 = 20 ways

Arrange A, B and C on selected position in 1 (For A at first of three selected positions) *2*1(for B and C) = 2 ways

Arrange remaining three on three remaining places in 3! = 6 ways

Total Favourable ways of arrangements of 6 horses = 6C3* (1*2*1)*(3!) = 20*2*6 = 240
Total Arrangements of 6 horses = 6! = 720

Probability = 240/720 =1/3

METHOD-3:
A has as much probability of being ahead of B and C as B has to be ahead of A and C as C has to be ahead of A and B therefore each one (of A, B and C) has equal probability to be winner among themselves

Hence probability of each horse (of A, B and C) to be ahead of other two = 1/3

Answer: option B