How many rectangles can be found in this? 
I saw there are 3 in each row, and 3 in each column hence,
$$9 \cdot 9 = 81$$
But the answer is $$441$$ somehow?
how do they get the answer?
 A: Since the figure in question contains $7$ vertical lines you can choose the two vertical edges of the rectangle in ${7\choose2}=21$ ways, and independently you can choose the two horizontal edges of the rectangle in $21$ ways.
A: A rectangle is defined by choosing a left-upper and a right-lower point. If we choose $(x,y)$ as a left upper point (using a grid notation starting at $(0,0)$ for left upper corner, going $+1$ for every square to the right or any square downwards), then the right lower point has $(u,v)$ to satisfy $u > x$ and $v > y$.
So for the $2\times 2$ we have $(0,0)$ as l-u, then we can have all $(x,y)$ with $x,y \in \{1,2\}$ as r-l point, so that gives 4 rectangles, using $(0,1)$ we have all $(x,y)$ with $x \in \{1,2\}, y \in \{2\}$, so 2 options, using $(1,0)$ we can use $(x,y), x > 1, y > 0$, so 2 options again, and with $(1,1)$ we only have $(2,2)$ as r-l point. So 4 + 2 + 2 +1 = 9 in total. Which confirms the statement.
Now for $6 \times 6$ we pick coordinates among $\{0,1,\ldots,6\}$. So 36 with $(0,0)$ as left upper, and generally $(6-j) \times (6-i)$ options when using $(i,j)$ as l-u point, where $i \le 5, j \le 5$, and you have to sum all those... Way more than 81...
A: Hint: $441=21\times21$, and
$$21=1+2+3+4+5+6$$
A: Number of squares here, = 6*6 + 5*5 + 4*4 + 3*3 + 2*2 + 1*1 = 91 squares.
Now try evaluating number of rectangles by considering cases like 2*1 rectangles, 3*2 rectangles, etc. 
Number of rectangles for a grid m*n is m(m+1)(n)(n+1)/4.
Therefore, it turns out to be 441.
Source: http://puzzlesland.com/mathprob/comb-subrect.html
