Answer: Option B
Two horses A and B, in a race of 6 horses . . . A has to finish before B.
If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4!
If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4!
If A finishes 3rd . . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4!
If A finishes 4th . . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways = 2 × 4!
If A finishes 5th . . . . B has to be 6th and the top 4 positions could be filled in 4! ways.
A cannot finish 6th, since he has to be ahead of B.
Therefore total number of ways:
= 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4!
= 120 + 96 + 72 + 48 + 24
= 360