To solve the problem of how many triangles can be formed with vertices at 18 points indicated on the perimeter of triangle $ABC$, we need to determine how many ways we can choose 3 points from these 18 points to form a triangle.

The number of ways to select 3 points from a set of 18 points is given by the combination formula:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

where $n$ is the total number of points, and $r$ is the number of points we want to choose.

In this case, $n = 18$ and $r = 3$. So, we calculate:

$\binom{18}{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = \frac{18 \times 17 \times 16}{6} = 816$

Thus, the number of triangles that can be formed is 816.

Would you like further details or have any questions?

Here are 5 related questions:

1. How many triangles can be formed from 10 points on a triangle’s perimeter?
2. What happens if 3 points are collinear? How would that affect the triangle count?
3. How does this solution generalize for polygons instead of triangles?
4. What is the formula to calculate combinations, and why is it used here?
5. How would the answer change if one or more points were inside the triangle?

Tip: When calculating combinations, remember that the order of selection does not matter, which is why we use combinations instead of permutations.