Let P_1,P_2… ,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≠ 15, is :

Given:${P}_{1},{P}_{2}\ldots ,{P}_{15}$P1​,P2​…,P15​ be $15$15 points on circle.

$\therefore$∴Total number of triangles$=C315$=C315

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

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

$i=2,j+k=13\Rightarrow (3,10),\ldots \ldots \ldots .,(6,7)=4$i=2,j+k=13⇒(3,10),……….,(6,7)=4 ways

$i=3,j+k=12\Rightarrow (4,8),(5,7)=2$i=3,j+k=12⇒(4,8),(5,7)=2 ways

$i=4,j+k=11\Rightarrow (5,6)=1$i=4,j+k=11⇒(5,6)=1 way

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

$\therefore$∴ The number of possible triangles using the vertices ${P}_{i},{P}_{j},{P}_{k}$Pi​,Pj​,Pk​ such that $i+j+k\neq 15$i+j+k=15 is equal to $C315-12=455-12=443$C315−12=455−12=443.