TestHub
TestHub

Mathematics - Permutation & Combination Question with Solution | TestHub

MathematicsPermutation & CombinationnPr & nCrHard2 minPYQ_2021
MathematicsHardsingle choice

LetP1, P2, P15be15points on a circle. The number of distinct triangles formed by pointsPi, Pj, Pksuch thati+j+k15,is :

Options:

Answer:
D
Solution:

Given:P1, P2, P15 be 15 points on circle.

Total number of triangles=C315

When ​​​​​​​i+j+k=15 then possible cases are

i=1, j+k=14(2, 12),(3, 11),(4, 10),(5, 9),(6, 8)=5 ways

i=2, j+k=13(3,10),.,(6,7)=4 ways

i=3, j+k=12(4 ,8),(5, 7)=2 ways

i=4, j+k=11(5, 6)=1 way

Hence, there are total 12 ways for i+j+k=15.

The number of possible triangles using the vertices Pi, Pj, Pk such that i+j+k15 is equal to C315-12=455-12=443.

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:nPr & nCr
2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...