The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. Practice Paper 1 Question 8

Given a grid of $$4\times4$$ points, how many triangles with their vertices on the grid can be drawn?

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Hints

• Hint 1
In how many ways may $$3$$ points be selected?
• Hint 2
When do $$3$$ vertices not form a triangle?
• Hint 3
Carefully consider which combinations of don't form triangles.

Solution

Triangles are formed by choosing any $$3$$ points that are not colinear. From a total of $$\binom{16}{3}$$ possible selected points, we exclude the combinations that form any straight lines:

• 10 lines pass through 4 points (4 horizontal, 4 vertical, 2 diagonals), hence $$10 \binom{4}{3}.$$
• 4 smaller diagonals passing through 3 points, hence $$4\cdot1.$$

In total we have $$\binom{16}{3}-10\binom{4}{3}-4 = 516.$$

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