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The pigeonhole principle is a simple yet powerful concept in combinatorial mathematics. It states that if you have more pigeons than pigeonholes and you attempt to place each pigeon into a hole, at least one hole will contain more than one pigeon. In mathematical terms, if you have more objects than containers and distribute the objects into the containers, at least one container will hold multiple objects.

Let's see how this principle applies to our problem. With 64 points on a plane and the constraint that we can only have 2001 distinct lines, we need to distribute the possible lines among these points.
This setting signifies that some points must share the same line to keep the total count down to or below 2001. Here, points sharing the same line are analogous to pigeons sharing a pigeonhole.
This guarantees that some lines need to accommodate multiple points to fit within the constraint, inevitably leading to several collinear points, at least four in this case.

Therefore, the pigeonhole principle helps us conclude that fewer lines than the maximum possible (2016 lines) force multiple points to share lines (collinearity), ensuring that at least four points are collinear.

Collinear points lie on the same straight line. If three or more points are collinear, they form a straight line when connected.

In our exercise, we have 64 points with a restriction that the maximum number of distinct lines that can be drawn is 2001.
Since the maximum possible number of lines between 64 points is 2016, this constraint implies the presence of collinear points to reduce distinct lines to 2001 or fewer.

If fewer than four points were collinear, each group of three points would lead to a higher number of combinations forming distinct lines, which exceeds the 2001 limit.
For example: \(\binom{64}{3} = \frac{64!}{3!(64-3)!} = \frac{64 \times 63 \times 62}{6} = 6944 \) forming a higher number of lines, far surpassing 2001.
Hence, the problem guarantees at least one set of four points sharing the same line to keep the distinct lines within the required count.

The combinations formula is an essential tool in combinatorial mathematics for determining how many ways you can choose a subset from a larger set without regard to order.
The general formula is given by: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) where \(n\) is the total number of items, and \(k\) is the size of each subset.

In our exercise, we use combinations to calculate the number of distinct lines possible from pairs of points: \(\binom{64}{2} = \frac{64!}{2!(64-2)!} = \frac{64\times 63}{2} = 2016\).
This shows the total number of lines that can theoretically be formed from 64 points.

Next, we compare this to the constraint of 2001 lines: Since 2016 lines exceed 2001, we infer the presence of collinear points.
To further solidify the restriction, calculating the number of lines from smaller subsets like groups of three: \(\binom{64}{3} = \frac{64!}{3!(64-3)!} = 6944\) explains why higher pairs and unnecessary distinct lines are avoided by ensuring collinear sets, solving the predetermined constraint effectively.
Thus, understanding combinations allows us to theorize and confirm the necessity of collinear points in combinatorial geometry problems.