Question

Find the number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the octagon .

Answer 1

Required number of triangles $=$total number of triangles$-$number of triangles having two sides common $-$number of triangles having one side common

${=}^8{C}_3-8-8\times 4$

$=\frac{8!}{3!5!}-8-32$

$=\frac{8\times 7\times 6\times 5!}{3\times 2\times 5!}-40$

$=56-40=16$