In a cycle race there are 5 persons named as J, K, L, M, N participated for 5…

Key insight: for any ordering of the five riders, either M is before N or N is before M. By symmetry, exactly half of the total permutations have M before N.

Total permutations of five distinct riders = 5! = 120. Therefore the number with M before N = 120/2 = 60.

Alternatively, count by M's finishing position:

• M is first: remaining 4 riders can be arranged in 4! = 24 ways.

• M is second: N must be in one of the 3 later positions (3 choices); the other 3 riders fill the remaining 3 spots in 3! = 6 ways, giving 3 × 6 = 18 ways.

• M is third: N can be fourth or fifth (2 choices), so 2 × 3! = 12 ways.

• M is fourth: N must be fifth, giving 1 × 3! = 6 ways.

• M is fifth: N cannot come after M, so 0 ways.

Summing: 24 + 18 + 12 + 6 = 60, which matches the symmetry result.