Q63.There are 5 points P1, P2, P3, P4, P5 on the side AB, excluding A and B, of a triangle ABC . Similarly there are 6 points P6, P7, … , P11 on the side BC and 7 points P12, P13, … , P18 on the side CA of the triangle. The number of triangles, that can be formed using the points P1, P2, … , P18 as vertices, is : (1) 776 (2) 796 (3) 751 (4) 771
What This Question Tests
This question assesses the understanding of combinations, specifically how to calculate the number of triangles that can be formed from a set of points, while accounting for collinear points.
Concepts Tested
Formulas Used
nCr = n! / (r! * (n-r)!)
📚 NCERT Sections This Tests
5.12 — Write All The Geometrical Isomers Of [Pt(Nh3)(Br)(Cl)(Py)] And How Many Of
Chemistry Class 11 · Chapter 5
5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
2.2 — A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 · Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
5.11 — Draw All The Isomers (Geometrical And Optical) Of:
Chemistry Class 11 · Chapter 5
5.11 Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2] + (ii) [Co(NH3)Cl(en)2] 2+ (iii) [Co(NH3)2Cl2(en)]+
📋 Question Details
- Chapter
- Permutation & Combination
- Topic
- Combination
- Year
- 2024
- Shift
- 04 Apr Shift 1
- Q Number
- Q63
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 7: Permutations and Combinations
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