Visual SolutionPYQ 2024 · Apr 4 Shift 1Tricky

Question

There are 5 points

P_1, P_2, P_3, P_4, P_5

on the side

AB

, excluding

A

and

B

, of a triangle

ABC

. Similarly there are 6 points

P_6, P_7, \ldots, P_{11}

on side

BC

and 7 points

P_{12}, P_{13}, \ldots, P_{18}

on side

CA

. The number of triangles that can be formed using the points

P_1, P_2, \ldots, P_{18}

as vertices, is:

(A)

776

(B)

796

(C)

751

(D)

771

Solution Path

{}^{18}C_{3} - {}^{5}C_{3} - {}^{6}C_{3} - {}^{7}C_{3} = 816 - 65 = 751

01Question Setup

1/4

5 points on AB, 6 on BC, 7 on CA (excluding vertices). Find number of triangles.

\text{Triangles} = \;?

02Total Combinations

2/4

{}^{18}C_{3} = 816

total ways. Subtract collinear triples on each side:

{}^{5}C_{3} + {}^{6}C_{3} + {}^{7}C_{3}

{}^{18}C_{3} = 816

03Subtract CollinearKEY INSIGHT

3/4

Collinear triples

= 10 + 20 + 35 = 65

. Triangles

= 816 - 65 = 751

816 - 65 = 751

04Final Answer

4/4

Number of triangles

= 751

\boxed{751}

Concepts from this question1 concepts unlocked

★

STANDARD

Count what you don't want and subtract from total: desired = total - undesired

|A| = |U| - |A^c|

When the constraint is complex, counting the complement is often far simpler. Saves 3-4 minutes on hard P&C problems.

At-least-one problemsDerangementsRestriction problems

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