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JEE Mains · Maths · STD 11 - 6. permutation and combination

There are \(5\) points \(\mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3, \mathrm{P}_4, \mathrm{P}_5\) on the side \(\mathrm{AB}\), excluding \(\mathrm{A}\) and \(\mathrm{B}\), of a triangle \(\mathrm{ABC}\). Similarly there are \(6\) points \(\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}\) on the side \(\mathrm{BC}\) and \(7\) points \(\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}\) on the side \(\mathrm{CA}\) of the triangle. The number of triangles, that can be formed using the points \(\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}\) as vertices, is :

  1. A  \(776\)
  2. B \(751\)
  3. C  \(796\)
  4. D \(771\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(751\)

Step-by-step Solution

Detailed explanation

\( { }^{18} \mathrm{C}_3-{ }^5 \mathrm{C}_3-{ }^6 \mathrm{C}_3-{ }^7 \mathrm{C}_3 \) \( =751\)
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