The first subtask can be solved by simply checking if there is at least one happy programmer for each possible assignment of tasks. There are ~N^N~ possible task assignments in total.

The second subtask can be solved using the inclusion-exclusion principle. If with ~a(S)~, where ~S \subseteq \{1, 2, \dots, N\}~, denotes the number of task assignments in which all programmers from set ~S~ are happy, then the number of good assignments is equal to

$$\displaystyle \sum_{S \subseteq \{1, 2, \dots, N\}} (-1)^{|S|+1} a(S)$$

Let ~S = \{s_1, s_2, \dots, s_k\}~. If ~s_1+s_2+\dots+s_k > N~ then obviously ~a(S) = 0~. Otherwise, it holds

$$\displaystyle a(S) = \binom{N}{s_1} \binom{N-s_1}{s_2} \dots \binom{N-(s_1+s_2+\dots+s_{k-1})}{s_k} (N-k)^{N-(s_1+s_2+\dots+s_k)}$$

Binomial coefficients can be precomputed using the well-known relation ~\binom{n}{k} = \binom{n-1}{k-1}+\binom{n-1}{k}~ in time complexity ~\mathcal O(N^2)~.

The total time complexity is ~\mathcal O(N 2^N)~.

Finally, the third subtask can be solved using dynamic programming. Let ~dp(n, k)~ be the number of assignments of ~k~ tasks to ~n~ programmers among which there is at least one happy programmer. There are two possibilities – we can give the ~n^\text{th}~ programmer exactly ~n~ tasks (and make him happy), or we won't. In the first case, if ~k \ge n~, the number of ways equals

$$\displaystyle \binom{k}{n} (n-1)^{k-n}$$

and in the other it equals

$$\displaystyle \sum_{0 \le i \le k, i \ne n} \binom{k}{i} dp(n-1, k-i)$$

The total time complexity is ~\mathcal O(N^3)~.