I need help with the following combinatorial problem. There are $ K $ passengers and $ K $ ships. The passengers are denoted by $ U_1, U_2, \dots, U_K $. The objective is to find in how many ways the $ K $ persons can board the ships, with the following conditions:
1) The ships have a maximum capacity of $ N $ passengers
2) There exists at least one ship with exactly $ N $ persons
3) The rest of ships can have an arbitrary number of persons that is less or equal to $ N $ (thus, other ships can have $ 0, 1, 2, \dots, N $ passengers).
4) Not all the ships need to be used.
What matters here is the number of ways in which the persons can be grouped with the constraints above (which ships are used is not relevant). For instance, one case is $ N = K $, when all the persons board one single ship. The remaining $ K-1 $ ships are left empty. Another case is when $ N = 1 $. Thus, each person exclusively boards a different ship (all ships have exactly one person because all the passengers are required to board).
I have one example with $ K = 4 $, where I will denote the passengers by U1, U2, U3 and U4.
Example 1: When $ K = 4 $, $ N = 1 $,
Combination 1: [U1] [U2] [U3] [U4] (the order does not matter).
Example 2: When $ K = 4 $, $ N = 2 $, we have
Combination 1: [U1 U2] [U3] [U4] [empty]
Combination 2: [U1 U3] [U2] [U4] [empty]
Combination 3: [U1 U4] [U2] [U3] [empty]
Combination 4: [U2 U3] [U1] [U4] [empty]
Combination 5: [U2 U4] [U1] [U3] [empty]
Combination 6: [U3 U4] [U1] [U2] [empty]
Combination 7: [U1 U2] [U3 U4] [empty] [empty]
Combination 8: [U1 U3] [U2 U4] [empty] [empty]
Combination 9: [U1 U4] [U2 U3] [empty] [empty]
Example 3: When $ K = 4 $, $ N = 3 $, we have
Combination 1: [U1 U2 U3] [U4] [empty] [empty]
Combination 2: [U1 U2 U4] [U3] [empty] [empty]
Combination 3: [U1 U3 U4] [U2] [empty] [empty]
Combination 4: [U2 U3 U4] [U1] [empty] [empty]
Example 4: When $ K = 4 $, $ N = 4 $, the only way is to put all the passengers in the same ship. Thus, we have
Combination 1: [U1 U2 U3 U4] [empty] [empty] [empty]
Update: Based on Paulo's answer, thus far I have this. In some cases this work but in others do not...
$$ \sum_{\substack{0 < n_1\leq\ldots\leq n_k\leq N\\[2pt]n_1+\ldots+n_k=K}} \frac{1}{(k-1)!} \frac{K!}{n_1!n_2!\cdots n_k!} $$
Useful results:
When $ K = 4 $, $ N = 1 $, the number of combinations is 1
When $ K = 4 $, $ N = 2 $, the number of combinations is 9
When $ K = 4 $, $ N = 3 $, the number of combinations is 4
When $ K = 4 $, $ N = 4 $, the number of combinations is 1
When $ K = 5 $, $ N = 1 $, the number of combinations is 1
When $ K = 5 $, $ N = 2 $, the number of combinations is 25
When $ K = 5 $, $ N = 3 $, the number of combinations is 20
When $ K = 5 $, $ N = 4 $, the number of combinations is 5
When $ K = 5 $, $ N = 5 $, the number of combinations is 1
One ship must have $N$ people, which is also the maximum number of people per ship (there are $K\choose N$ ways of doing this). After that, this problem reduces to distributing $K-N$ people by $K-1$ ships, where the maximum number of people per ship is $N$.
The number of ways of dividing $n$ people in one group of $n_1$ people, one group of $n_2$, one of $n_3,\ldots,$ and one of $n_k$ can be shown to be
$$\frac{n!}{n_1!n_2!\cdots n_k!}$$
And now you consider every possible way of choosing how many elements are in each group and you sum all of them, and multiply by the number of ways you can choose the $N$ people for the "fixed" ship: $${K\choose N} \left(\sum_{k=1}^{K-1}\sum_{\substack{0< n_1\leq\ldots\leq n_k\leq N\\[2pt]n_1+\ldots+n_k=K-N}}\frac{\left(K-N\right)!}{n_1!n_2!\cdots n_k!}\right)$$
which I believe to be the answer, but I don't know whether it can be simplified.
I hope this helps.