JEE Mains · Maths · STD 11 - 6. permutation and combination
Let ABC be a triangle. Consider four points \(p _1, p _2\), \(p _3, p _4\) on the side AB , five points \(p _5, p _6, p _7, p _8, p _9\) on the side BC and four points \(p _{10}, p _{11}, p _{12}, p _{13}\) on the side AC . None of these points is a vertex of the triangle ABC . Then the total number of pentagons, that can be formed by taking all the vertices from the points \(p _1, p _2, \ldots . p _{13}\), is ___ .
- A 640
- B 650
- C 660
- D 670
Answer & Solution
Correct Answer
(C) 660
Step-by-step Solution
Detailed explanation
Case 1: 2 from \(AB , 2\) from \(BC , 1\) from AC \(\binom{4}{2} \cdot\binom{5}{2} \cdot\binom{4}{1}=6 \cdot 10 \cdot 4=240\) Case 2 : 2 from \(AB , 1\) from \(BC , 2\) from AC \(\binom{4}{2} \cdot\binom{5}{1} \cdot\binom{4}{2}=6 \cdot 5 \cdot 6=180\) Case 3 : 1 from \(AB , 2\)…
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