Number of Triangles that Can Be Formed from 8 Points

We are given 8 points in total, with the following conditions:

• Two lines $p$ and $q$ have 3 points each.
• Point $r$ lies on both lines $p$ and $q$, meaning $r$ is a common point between the two lines.
• There are 4 other distinct points, which means there are 8 total points: 3 points on line $p$, 3 points on line $q$, and 2 other points not on either line.

Step 1: Triangles from 8 points without constraints

The total number of triangles that can be formed from 8 points (without considering any constraints) is simply the combination of choosing 3 points out of 8. This can be calculated as: $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Step 2: Subtract degenerate cases (collinear points)

Triangles cannot be formed from 3 collinear points, so we need to exclude the degenerate cases.

• Line $p$ has 3 points. A triangle cannot be formed by selecting all 3 points from line $p$, so we subtract this case: $\binom{3}{3} = 1 \text{ (degenerate case for line } p\text{)}$

• Similarly, line $q$ also has 3 points, so we subtract this case: $\binom{3}{3} = 1 \text{ (degenerate case for line } q\text{)}$

Now, subtract these degenerate cases: $56 - 1 - 1 = 54$

So, 54 triangles can be formed from these 8 points.

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Number of Quadrilaterals that Can Be Formed from 8 Points

To form a quadrilateral, we need to choose 4 points from the 8 total points. This can be calculated as:

$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$

Since quadrilaterals cannot be degenerate (no 4 points are collinear on any of the lines), there are no degenerate cases to subtract in this situation.

So, the total number of quadrilaterals that can be formed is 70.

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Summary:

• Triangles: 54
• Quadrilaterals: 70

Would you like any details or have further questions?

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Related Questions:

1. What is the general formula for calculating combinations?
2. How can we calculate the number of degenerate polygons in a set of points?
3. What happens if more points are collinear in the above problem?
4. How does the presence of overlapping points affect polygon formation?
5. Can we use similar techniques to count other polygons, like pentagons?

Tip:

When solving combinatorics problems, always consider if there are degenerate cases (like collinear points) that need to be subtracted from the total combinations.