Partition an integer $n$ into exactly $k$ distinct parts

If you partition $n$ into exactly $k$ distinct parts, then the $i$-th smallest part must be at least $i$, therefore strictly larger than $i-1$. So you can start reserving $i-1$ units for the $i$-th smallest part, which amount to a total of $\binom k2$ reserved units. Then (assuming $n\geq\binom{k+1}2$ without which no solution is of course possible) you can find a partition of $n-\binom k2$ into exactly $k$ parts, and then add the $i-1$ reserved units to the $i$-th smallest part of that partition (it may be ex-aequo with another part, but after these additions it will be distinct from any other parts, and exactly $i$-th smallest among them). This gives every partition of $n$ into exactly $k$ distinct parts exactly once.

The reason for not reserving $i$ parts is than you would then need a partition of $n-\binom{k+1}2$ into at most $k$ parts, and this case did not occur in the list of cases you said you knew how to handle. But in any case, the number of ways to partition any $m$ into at most $k$ parts is equal to the number of ways to partition $m+k$ into exactly $k$ parts, by a similar argument (reserve one unit for each part).