| If \(N > 2K\), then the answer is "`NO`", because we lack the capacity to hold all the parts. | |
| If there are \(3\) parts of the same style, then the answer is "`NO`". By the pigeonhole principle, any assignment of parts to cases will yield a case with two or more parts of the same style. | |
| Otherwise the answer is "`YES`" using the following strategy: First, for any style we have two parts of, put one in each case. Next, evenly distribute the remaining parts (each being a unique style in the store) between the two cases. Informally, this distributes the parts as close to evenly as possible, so we will have enough room in each case. | |
| Formally, one case will have \(\lfloor N/2 \rfloor \) parts and the other will have \(\lceil N/2 \rceil \) parts. Since \(N \le 2K\), we have \(N/2 \le K\), and thus \( \lceil N/2 \rceil \le \lceil K \rceil = K\) (inequality preserved since \(\lceil x \rceil\) is a nondecreasing function). Having shown \(\lfloor N/2 \rfloor \le \lceil N/2 \rceil \le K\), we see that neither case will exceed capacity. | |
| [See David Harmeyer's solution video here.](https://youtu.be/8Eg2-HIUP-w) | |