Using diagonals from a common vertex, how many triangles could be formed from a 21-gon?

To determine how many triangles can be formed from a 21-gon (a polygon with 21 sides), we can use a combinatorial approach. 

The key here is that we need to select three vertices from the 21-gon to form a triangle. The diagonals will connect vertices, but for each triangle, we don't care whether the edges are sides of the polygon or diagonals; we just need to select 3 distinct vertices.

The number of ways to choose 3 vertices from 21 vertices (to form a triangle) is given by the combination formula:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

Where:
- \( n = 21 \) (the number of vertices in the 21-gon)
- \( k = 3 \) (the number of vertices we need to choose to form a triangle)

So we compute:

\[
\binom{21}{3} = \frac{21!}{3!(21-3)!} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 1330
\]

Thus, **1330** triangles can be formed from a 21-gon using its diagonals and sides.

Let me know if you'd like further clarification or a breakdown of this calculation!

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Here are some related questions to expand on this:
1. How does the formula for combinations change if we are choosing different numbers of vertices (e.g., choosing 4 vertices)?
2. What is the maximum number of diagonals that can be drawn in a 21-gon?
3. How many diagonals are there in a polygon with \( n \) sides?
4. How would the number of triangles change if the polygon had 22 sides instead of 21?
5. What is the relationship between the number of diagonals and the number of triangles in a polygon?

**Tip:** Remember that the formula for combinations is always useful for counting ways to choose distinct items, like vertices, without regard to order!

Learn how to calculate the number of triangles that can be formed from a 21-gon using its diagonals. This problem involves using the combination formula to select 3 vertices from the 21 vertices of the polygon. The result is 1330 triangles, suitable for students in Grades 9-11.

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