If $PQRS$ is a convex quadrilateral with. 3, 4, 5 and 6 points marked on sides $PQ, QR, RS$ and $PS$ respectively, then the number of triangles with vertices on different sides is

Clearly,

Total number of triangles

= Number of triangles with vertices on sides$PQ, QR$ and $RS$

+ Number of triangles with vertices on sides $QR, RS$ and $PS$

+ Number of triangles with vertices on sides $RS, PS$and $PQ$

+ Number of triangles with vertices on sides$PS, PQ$ and $QR$

$\begin{array}{l}{=}^3{C}_1{\times }^4{C}_1{\times }^5{C}_1{+}^4{C}_1{\times }^5{C}_1{\times }^6{C}_1{+}^5{C}_1{\times }^6{C}_1{\times }^3{C}_1\\ {+}^6{C}_1{\times }^3{C}_1{\times }^4{C}_1\end{array}$

$= 60 + 120 + 90 + 72 = 342.$