permutation

Md.Nazrul Islam wrote:A polygon has 44 diagonals .Number of sides are .

We can also solve this without using combinations.

If there are n sides, there are n vertices.

Let's examine 1 vertex. From this vertex, how many other vertices can this vertex connect to to create a diagonal line? Well, we can't draw a diagonal line to the 2 adjacent vertices, and we can't draw a line to the vertex we are starting from.
So, from one particular vertex, there are n-3 candidates to create a diagonal line to.

So, we have n vertices altogether, and from each vertex we can create n-3 diagonals.
This means that the total number of diagonals = n(n-3)
Now keep in mind that this happens to count each diagonal twice.
So, to account for this duplication, we can say that:

In any n-sided convex polygon, the total number of diagonals = n(n-3)/2

Now onto the question. We're told that there are 44 diagonals.

So, n(n-3)/2 = 44
Multiply both sides by 2: n(n-3) = 88
Expand: n^2 - 3n = 88
Set quadratic equal to zero: n^2 - 3n - 88 = 0
Factor: (n+8)(n-11)=0
So, n=-8 or n=11
n can't be negative, so the answer is [spoiler]n=11[/spoiler]

Cheers,
Brent