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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\$.

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 \$\$inom32inom52\$\$

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 + 12inomn + 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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