Is it 60 ?

I believe this was discussed in a recent thread. Total number of
combinations = 5! = 120. Out of this, half the number of times, M will
be before B and half the number of times B will be before M.

So, total = 120 / 2 = 60.

Alternatively, you can look at positions to compute this

When M is in first place, B can finish in 4 different places (i.e. MB---, M-B--
etc.). For each such case, we have 3! permutations for the other set of 3
people to finish the race.

So, when M is in first place, we have 4 * 3! possibilities.

Similarly, computing for M in second place, third place etc. we get

3! * (4 + 3 + 2 + 1) = 6 * 10 = 60 possibilities

Thank you Jay. Yes, the answer is 60.

Jay's method is the best...Still

Select 2 position from the total positions = 5C2

Assume that in the above position always Meg is ahead of the other

and the rest position can be filled in 3*2 ways

so total no of ways in which Meg finished before Bob

5C2 * 3* 2

and hence = 60