Intersecting parellelograms!

$^{15}C_{2} \times ^{16}C_{2} =12600$

We might as well think of squares formed by n horizontal lines and m vertical lines.

Each square is defined by a vertex and the diagonally opposite vertex.

These lines intersect at $m\times n$ nodes which are the potential vertices of the squares. Choose one of these to be the first vertex of the square. This can be done in $n\times m$ ways. The second vertex cannot be on either of the two lines passing through the first vertex, so it can be chosen in $(n-1)(m-1)$ ways.

The number of ways both vertices can be chosen is thus

$nm(n-1)(m-1)$

However this formula quadruple counts the number of squares because any one of the four vertices could have been chosen as the first vertex!

The number of squares is thus $P=\dfrac{nm(n-1)(m-1)}{4}$

When n=15 and m=16 we thus find

$\dfrac{P}{100}=\dfrac{15\times 16 \times 14 \times 15}{400}=\boxed{126}$

If we consider 2 horizontal parallel lines intersecting with 2 vertical parallel lines, the number of parallelograms formed is $1$ , i.e.

2 horizontal and 3 vertical would give us $3$ parallelograms, i.e.

3 horizontal and 3 vertical would give us $9$ parallelograms, i.e.

... and so on.

So for $h$ horizontal and $v$ vertical lines, the number of parallelograms formed is

$\sum_{a=1}^{h-1}\sum_{b=1}^{v-1}(a\times b)=\sum_{a=1}^{h-1}a\times\sum_{b=1}^{v-1} b$

Now $\sum_{x=1}^m x=\frac{m(m+1)}{2}$ ,

so for $h=15$ and $v=16$ , the number of parallelograms formed is

$P=\sum_{a=1}^{14} a\times\sum_{b=1}^{15} b=\frac{14(15)}{2} \times \frac{15(16)}{2}=12600$

Therefore $\frac{P}{100}=\boxed{126}$

I drew figures for 2x2 , 2x3 , 3x3 , 3x4 , 4x4 , 4x5 , and 5x5 grids and counted the rectangles. I found, respectively, 1, 3, 9, 18, 36, 60, and 100.

I noticed that every other grid was a perfect square, and the grid between each pair of perfect squares was the geometric mean of those perfect squares.

The 2x2 grid had 1^2 parallelogram, the 3x3 grid had 3^2 parallelograms, the 4x4 grid had 6^2 parallelograms, and the 5x5 grid had 10^2 parallelograms. Continuing this pattern, I determined that the 15x15 grid would have 105^2 parallelograms and the 16x16 grid would have 120^2.

The geometric mean of 105^2 and 120^2 is 12600. Divided by 100 gives you 126.

this is same as finding no. of rectangles of any size in a rectangular chess board of 14 x 15

Which is done by [14 (14+1)/2] [15*(15+1)/2]=12600