How To Really Solve This Combinatorics Puzzle?

This combinatorics puzzle offers you a nice opportunity to brush up on your logical reasoning skills. You are presented with a total of thirteen planar points. Of the thirteen, the six pink points are collinear. That is, they lie along a straight line (visualized as the pink broken line). Similarly, the four green points and the three black points are collinear along the green broken line and the black broken line respectively.

Given this premise, your job is to figure out how many straight lines you can construct using these planar points. Before you proceed, note that we are talking about straight lines here and not line-segments. Unlike a line-segment, a straight line does not have end points; it extends indefinitely either side.

Here are the conditions for the puzzle:

1. Each line should use at least two points.

2. If two or more lines are collinear, they should be considered as one line. In other words, if there are an infinite number of lines possible through 2 or more collinear points, only one of them should be considered for the final count.

Hint:

You could try taking advantage of certain geometrical properties of straight lines. Using combinatorics, you would then be able to map your general understanding onto the given problem space.

Spoiler Alert:

If you wish to try this puzzle out on your own, now is the time. Once you’ve given the puzzle a try, you may choose to continue reading this essay. Beyond this section, I will be explicitly discussing solutions to the puzzle.

This essay is supported by Generatebg

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Solving the Combinatorics Puzzle for a Special Case

The first thing we need to note in this puzzle is that the problem space consists of three different sets of collinear points. This adds an annoying layer of complexity to the puzzle.

Instead of worrying about collinear points right from the start, let us make it a tad bit simpler by looking at a hypothetical problem space in which no three points are collinear. Such a space of thirteen points would look like this:

Now, we just need to worry about two factors:

1. The total number of points in the plane

2. The total number of possible lines that can be constructed using these points.

The fundamental property of a line is that it must pass through at least two points. In our special problem case, no three points are collinear. What this means is that we could construct a line by just letting it pass through any two points randomly.

The question then becomes:

“How many combinations of point-pairs could be chosen from a total of thirteen planar points?”

The answer is: 13C₂ = 78 (where C denotes the combination function from combinatorics).

Now that we have arrived at a solution for our special case, let us try and extend this result to our original problem space.

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How to Solve the Combinatorics Puzzle?

The special case that we just solved had no more than two collinear points. But in our original problem, we have three different sets of collinear points. This means that our original problem represents a subset of the special case.

In other words, the total number of possible straight lines in our original problem space has to be necessarily less than that in the special problem space. So, we need to subtract something from the special case’s solution (78). What could that be?

Subtracting Redundant Combinations

Let us take a look at the set of pink collinear points. If I construct a line by allowing it to pass through any two of these points, the line also passes through all the rest of the points. So, one line covers all combinations as far as this collinear set of points is concerned.

By analogy, we could arrive at the same conclusion for the green and black collinear sets of points respectively. By doing this, we arrive at an equation where we subtract the individual collinear line combinations from the total number of all possible line combinations for the special case. We can express this equation as follows:

13C₂–6C₂–4C₂–3C₂

Before we call it a day, there is one last thing we need to take into account. By subtracting the above terms, we have discounted all possible collinear combinations. This is not true. We have three sets of collinear points. So, we will need to take the corresponding three collinear lines into account as well. If we do this, we arrive at the following equation:

13C₂–6C₂–4C₂–3C₂ + 3

= 78–15–6–3+3

=57

There we go. We can construct a total of 57 straight lines in the original combinatorics puzzle space!

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Further reading that might interest you: How Much String Would You Need To Wrap The Earth? and Are We Living In A Simulation?

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