First of all we can assume k >= m since otherwise we have a 100% chance that one server fails.

Next how many options do the first job have? Its k. The second job have k-1 until the mth job has k-m+1 possibilities.

So this yields k!/(k-m)! Diffrent possibilieties to assign the jobs without fail.

In total we have k^m possibilieties to asign the jobs in total.

This yield a probability of k!/((k-m)!k^(m))