How many rectangles with the dark dot are there?

How many rectangles with the dark dot are there?

I know how to get the total number of rectangles using combinations, but how can I find the number of rectangles containing the dark dot?

Is this answer correct?

$3*2*4*2 = 48$

Solution 1:

Lets call the vertical lines as $v_1,v_2,...,v_7$ from left to right, and call the horizontal lines as $h_1 ,h_2,h_3,h_4$ from top to bottom.

So we should select one line form each of the following sets $S_1=\{v_1,v_2,v_3\}$, $S_2=\{v_4,v_5,v_6,v_7\}$, $S_3=\{h_1,h_2\}$, $S_4=\{h_3,h_4\}$ to construct rectangles containing black dot. For example one of the possible rectangles is composed of $v_2,v_5$ and $h_1,h_4$.

So $$\binom{3}{1}\binom{4}{1}\binom{2}{1}\binom{2}{1}=3\times 4 \times 2 \times 2 =48$$

Your answer is correct!!

Solution 2:

Following the hint of @Angelica:

A rectangle is uniquely determined if we know its left upper corner and its right lower corner. If the black dot is contained in the rectangle then the left upper corner is the intersection of one of the $2$ horizontal grid lines that lie above the black dot and one of the $3$ the vertical grid lines that lie left of the black dot. There are $2\times3=6$ such intersections. The right lower corner is the intersection of one of the $2$ horizontal grid lines that lie below the black dot and one of the $4$ vertical grid lines that lie right to the black dot. There are $2\times 4=8$ such intersections. The left upper corner and the left loser corner can be selected independently and each selection gives a different rectangle.

So we have $6\times 8$ different rectangles that contain the black dot.