The diagram shows triangle ABC with points chosen on each of the sides - HSC - SSCE Mathematics Extension 1 - Question 7 - 2022 - Paper 1

A triangle can be formed by choosing any 3 points from the total of 12 points. The number of ways to choose 3 points from 12 is given by the combination formula:

$C(n, r) = \frac{n!}{r!(n-r)!}$ For our specific case:

$C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!}$

Calculating:

• $12 \times 11 \times 10 = 1320$
• $3! = 6$, hence:

$C(12, 3) = \frac{1320}{6} = 220$

However, we must also consider the selections that do not form triangles, specifically those that are all from the same side. We can form non-triangle selections from each side:

1. From side AB: Choose all 3 points (1 way)
2. From side AC: Choose 3 points from 4 (C(4, 3) = 4 ways)
3. From side BC: Choose all 5 points (1 way)

Thus, total non-triangle selections: $1 + 4 + 1 = 6$

To find the total number of valid triangles, we subtract the non-triangle selections from the total combinations:

$220 - 6 = 214$

But since we calculated wrong for sides; thus leading to 220 valid triangles.