There are 5 points P1, P2, P3, P4, P5 on the side AB, excluding A… — Solution | JEE Mathematics

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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 the side $BC$ and $7$ points $P_{12}, P_{13}, \ldots, P_{18}$ on the side $CA$ of the triangle. The number of triangles that can be formed using the points $P_1, P_2, \ldots, P_{18}$ as vertices is:

A
$776$
B
$751$
C
$796$
D
$771$

Answer

Correct answer:B

Step-by-step solution

Counting non-collinear triples

Given: There are $18$ points on the sides of triangle $ABC$ : $5$ on $AB$ , $6$ on $BC$ , and $7$ on $CA$ .

Find: The number of triangles formed using these $18$ points as vertices.

Any triangle is formed by choosing $3$ non-collinear points.

The total number of ways to choose $3$ points from $18$ is

\binom{18}{3} = 816

Now subtract the collinear selections. Since points lying on the same side are collinear, the number of such cases is

\binom{5}{3} + \binom{6}{3} + \binom{7}{3} = 10 + 20 + 35 = 65

Therefore, the number of triangles is

816 - 65 = 751

Therefore, the correct option is B.

Common mistakes

Counting all $\binom{18}{3}$ selections as triangles. This is wrong because three points on the same side are collinear and do not form a triangle. Subtract the collinear cases from the total.
Using $\binom{5}{2} + \binom{6}{2} + \binom{7}{2}$ for invalid cases. This is wrong because a degenerate triangle requires $3$ collinear points, not $2$ . Use $\binom{5}{3}, \binom{6}{3}, \binom{7}{3}$ instead.
Including points $A, B, C$ in the count. This is wrong because the question explicitly says only $P_1$ to $P_{18}$ are used as vertices. Count only those given points.

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Answer

Step-by-step solution

Common mistakes

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