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In a $2 × 4$ rectangle grid shown below, each cell is a rectangle. How countless rectangles deserve to be observed in the grid?

### My attempt:

I uncovered a formula somewhere,

Number that rectangles space $= m(m+1)n(n+1)/4 = 2 imes4 imes3 imes5/4 = 30$.

Can you please describe this?

### Upadte -

The postHow countless rectangles or triangles. Looks similar, however that has $3 imes 4$ grid and also I need more variant explanation.

To type a rectangle, we must choose two horizontal sides and also two vertical sides. Due to the fact that there room three horizontal lines, us can choose the horizontal sides in $inom32$ ways. Because there are 5 vertical lines, we can select the vertical political parties in $inom52$ ways. The variety of rectangles us can kind is $$inom32inom52$$

In general, the number of rectangles have the right to be formed in a $m imes n$ rectangular grid with $m + 1$ horizontal lines and also $n + 1$ vertical lines is the variety of ways us can select two of the $m + 1$ horizontal lines and also two of the $n + 1$ upright lines to be the sides of the rectangle, which is $$inomm + 12inomn + 12 = frac(m + 1)!(m - 1)!2! cdot frac(n + 1)!(n - 1)!2! = frac(m + 1)m2 cdot frac(n + 1)n2$$

## Not the prize you're spring for? Browse various other questions tagged combinatorics discrete-sommos.netematics permutations combinations or ask your very own question.

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