Just require an easy formula!

No. of rectangles in a chess board of $n \times m$ = $\large \binom{n+1}{2} \times \binom{m+1}{2}$

Note : $\large \binom{a}{b}$ = $\dfrac{a!}{b! \cdot (a-b)!}$

We can think of it in this way: A 5 * 5 chessboard will have 6 vertical lines and 6 horizontal lines. A rectangle is made by any 2 horizontal lines and any 2 vertical lines . Hence no. of rectangles possible here is 21C2 * 15C2