CUET UG 2025 — General Test Quantitative Reasoning

Total points in the plane = 9

Points that lie on the same line = 4 (collinear)

Remaining points = 5 (not on that line)

To form a triangle, 3 points are needed that are not all on the same line.

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Total ways to select any 3 points from 9 points:

$$C(9,3) = \dfrac{9!}{3! \times 6!}$$C(9,3)=3!×6!9!​

$$C(9,3) = \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}$$C(9,3)=3×2×19×8×7​

$$C(9,3) = \dfrac{504}{6}$$C(9,3)=6504​

$$C(9,3) = 84$$C(9,3)=84

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If 3 points are selected from the 4 collinear points, they will all be on the same line and will not form a triangle.

Number of ways to pick 3 points from the 4 collinear points:

$$C(4,3) = \dfrac{4!}{3! \times 1!}$$C(4,3)=3!×1!4!​

$$C(4,3) = \dfrac{4}{1}$$C(4,3)=14​

$$C(4,3) = 4$$C(4,3)=4

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Number of triangles = Total combinations - Combinations that don't form triangles

Number of triangles $= 84 - 4$=84−4

Number of triangles $= 80$=80

Therefore, the number of triangles that can be formed is 80.