Ten points are marked on a circle. The number of distinct convex polygons of three or more sides that can be drawn using some or all of the ten points as vertices is:

Three vertices are chosen randomly from the seven vertices of a regular $7$-sided polygon. The probability that they form the vertices of an isosceles triangle is

There are $n$ distinct points on a circle. If the number of pentagons that can be formed using these points as vertices is equal to the number of triangles that can be formed,then the value of $n$ is:

Six '$X$'s have to be placed in the squares of the figure such that each row contains at least one '$X$'. In how many different ways can this be done?

Consider a rectangle $ABCD$ having $5, 6, 7, 9$ points in the interior of the line segments $AB, BC, CD, DA$ respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta-\alpha)$ is equal to:

If three of the six vertices of a regular hexagon are chosen at random,then the probability that the triangle formed with these chosen vertices is equilateral is