# Knapsack with fixed size

We have $$3n$$ items with profits $$p_1, \dots, p_{3n}$$ (sum = $$P$$) and weights $$w_1,\dots,w_{3n}$$ (sum = $$W$$). We want to determine whether we can choose exactly $$n$$ items with profit at least $$P/2$$ and weight at most $$W/2$$.

Is this problem NP-hard? It is very similar to the usual knapsack problem, but the constraint of choosing exactly $$n$$ items makes it difficult to reduce directly from knapsack.

The problem is NP-Hard. You can reduce from the partition problem: given a multi-set $$X = \{x_1, \dots, x_n\}$$ of non-negative integers, is there a subset $$S$$ of $$X$$ such that $$\sum_{x \in S} x = \frac{1}{2} \sum_{x \in X} x$$?
To obtain an instance of your problem you can create one $$y_i$$ for each integer $$x_i$$ and set $$p_i = w_i = x_i$$. Then create $$2n$$ additional "dummy" items with weight 0 and profit 0.
If there is a solution $$S = \{x_{i_1}, \dots, x_{i_k}\}$$ to the partition instance, then you can obtain a solution to your problem by selecting the items $$y_{i_1}, \dots, y_{i_k}$$ plus $$n-k$$ dummy items.
If there is a solution to the instance of your problem with non-dummy items $$y_{i_1}, \dots, y_{i_k}$$ (and any number of dummy items) both the total weight and the total profit of the selected items must be exactly $$\frac{1}{2} \sum_{x \in X} x$$. Then, the set $$S = \{x_{i_1}, \dots, x_{i_k}\}$$ is a solution for the partition instance.