Given:$P_1, P_2\dots , P_{15}$ be $15$ points on circle.

$\therefore$Total number of triangles$={}^{15}C_3$

When $\mathrm{i}+j+k=15$ then possible cases are

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

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

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

$i=4, j+k=11\Rightarrow (5, 6)=1$ way

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

$\therefore$ The number of possible triangles using the vertices $P_i, P_j, P_k$ such that $i+j+k\ne 15$ is equal to ${}^{15}C_3-12=455-12=443$.