The question asks: Suppose there are 1350 people seated in a row of 2021 chairs. Prove that there are 3 consecutive non-empty (taken) chairs.
Here was my logic: Assume (by contradiction) that the there are no 3 consecutively taken chairs. So the most taken chairs we can have consecutively is 2. Now, the seating that will satisfy this whilst taking up the least amount of "space" is placing an empty chair between every pair of people. Something like this, where A is a person sitting in a chair and each _ is an empty chair: AA_AA_AA_AA_ and so on. Now, there are 675 pairs of people, and each pair (except for the last one) needs to have an empty chair following them. This seating arrangement would require 674*2 + 674 = 2022 chairs (the number of people + the number of empty chairs) which is more than we have.
Contradiction. So there must be at least one instance of 3 consecutive non-empty chairs. In another line of thinking, by the pigeonhole principle, 1 person must "share" with an empty chair and therefore fill one of the empty chairs separating 2 pairs of people, which leads to 3 consecutive non-empty chairs.
Is this correct? I submitted this solution for a math class and received a very low grade, as if my reasoning were completely off. I'd like to know if I'm missing something.