JEE Main 2021 (Online) 1st September Evening ShiftQuestion 1
Let $P_1,P_2,\ldots,P_{15}$ be $15$ points on a circle. The number of distinct triangles formed by points $P_i,P_j,P_k$ such that $i+j+k\neq15$, is :
Permutation and Combination
Solution
Concept. No three points of a circle are collinear, so every 3 points give a triangle. Remove those with $i+j+k=15$. Formulas used. (1) $\binom{15}3$ triangles in allStep 1 - All. $\binom{15}3=455$. Step 2 - $i<j<k$, sum 15. $i=1$: $(2,12),(3,11),(4,10),(5,9),(6,8)$: 5 $i=2$: $(3,10),(4,9),(5,8),(6,7)$: 4 $i=3$: $(4,8),(5,7)$: 2 $i=4$: $(5,6)$: 1 Total $12$. Step 3 - Answer. $455-12=443$. Answer: D