Find the number of diagonals in the convex polygon of n sides .

Explanation

Diagonal of the polygon is formed, if two non-consecutive vertices of polygon are joined.
So, number of diagonals=number of ways we can select two non-consecutive vertices of polygon.
The first vertex can be selected in n ways.
Let `A_(1)` is chosen first.
Now diagonal cannot be formed if any of `A_(2) " and" A_(n)` is chosen.
Hence for `A_(1)` another vertex can be selected in (n-3) ways from remaining (n-3) vertices.
Similarly, there are (n-3) diagonals passing through each vertex.
So, by multiplication principle of counting,
Number of diagonals `=n xx (n-3)`
But, in above answer there is double counting.
Let `A_(1)` is chosen as first vertex, then against it sometimes `A_(4)` is chosen as second vertex.
Similarly, when `A_(4)` is chosen as first vertex, then against it sometimes `A_(1)` is chosen as second vertex.
Thus, each pair of vertices is selected twice.
Hence total number of diagonals `=(n(n-3))/(2)`