CareerCup

This is equal to 1-(probability that they all carry different number of passengers). That probability in paranthesis is equal to (m+1)!/((m+1-n)! * (m+1)^n). This is given that (m+1) >= n. If (m+1) < n, then the probability is 1.

- m+1 because maximum m passengers mean m+1 different possiblities (a bus could carry 0 passengers).
- The first bus has (m+1) possibilities in terms of number of passengers. The second has m+1, the third has m+1 and so on. Total number of orderings possible is (m+1) ^ n.
- For the number of passengers to be different in each bus, first bus has m+1 possibilities, second has m, third has m-1 nd so on. For n buses, this is equal to (m+1)!/(m+1-n)!. Dividing this by (m+1)/n gives the possibility. Subtract this number from one to find at least two buses with the same amount of passengers.

should we assume that probability of a bus carrying 0,1,2,3,.. m are all equal

1-(((m+1)Cn)n!/(m+1)^n)

The correct answer should be:
1 - ( (m+1)! / (m-n-1)! ) / ( (m+1)^n )

This is what I think:

1 - P (all the buses having distinct numbers of passengers)

Which boils down to:

1 - m * (m - 1) * ... * (m - (n - 1))

(you can derive this with principle of counting)

Which is:

1 - m! / (m - n)!

Now, an important assumption here is that

m > n

otherwise the answer is just

1

, because no matter how you allocate passengers to buses, at least two of the buses will have equal passengers.

Please correct me if I'm wrong.