Out of $n$ points in a plane,$p$ points are collinear. (No three of the remaining points are collinear). The number of lines that can be drawn passing through these points is:

$A$ polygon has $54$ diagonals. The number of sides of this polygon is

Out of $20$ points,no three points are collinear except for $4$ points which are collinear. How many lines can be drawn by joining these points?

In the given figure,$6$ '$A$'s should be written in such a manner that every row contains at least one '$A$'. The number of ways this is possible is

There are $5$ points $P_1, P_2, P_3, P_4, P_5$ on the side $AB$,excluding $A$ and $B$,of a triangle $ABC$. Similarly,there are $6$ points $P_6, P_7, \ldots, P_{11}$ on the side $BC$ and $7$ points $P_{12}, P_{13}, \ldots, P_{18}$ on the side $CA$ of the triangle. The number of triangles that can be formed using the points $P_1, P_2, \ldots, P_{18}$ as vertices is:

Let $ABC$ be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side $AB$, five points $p_5, p_6, p_7, p_8, p_9$ on the side $BC$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side $AC$. None of these points is a vertex of the triangle $ABC$. The total number of pentagons that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$ is . . . . . . .