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 , … , P 11 on the side BC and 7 points P 12 , P 13 , … , P 18 on the side CA of the triangle. The number of

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Solution

Q.
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 $P_{6}, P_{7}, \dots, P_{11}$ on the side BC and 7 points $P_{12}, P_{13}, \dots, P_{18}$ on the side CA of the triangle. The number of triangles, that can be formed using the points $P_{1}, P_{2}, \dots, P_{18}$ as vertices, is: [2024]

1 796

2 751

3 771

4 776

Ans.
(2)

Number of points on side AB = 5
Number of points on side BC = 6
Number of points on side AC = 7

Number of ways of selecting three points from side AB = ${}^{5}{C_{3}} $

Number of ways of selecting three points from side BC = ${}^{6}{C_{3}} $

Number of ways of selecting three points from side AC = ${}^{7}{C_{3}}$

Total number of triangles that can be formed using the points $P_{1}, P_{2}, \dots, P_{18} = {}^{18}{C_{3}} - ({}^{5}{C_{3} + {}^{6}{C_{3} + {}^{7}{C_{3}}}})$

$= {}^{18}{C_{3} - {}^{5}{C_{3}} - {}^{6}{C_{3}} - {}^{7}C_{3}} = 816 - 10 - 20 - 35 = 751$

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