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In a $2 × 4$ rectangle grid shown below, every cell is a rectangle. How countless rectangles have the right to be it was observed in the grid? ### My attempt:

I uncovered a formula somewhere,

Number the rectangles are $= m(m+1)n(n+1)/4 = 2\times4\times3\times5/4 = 30$.

The postHow plenty of rectangles or triangles. Look at similar, however that has actually $3 \times 4$ grid and also I need an ext variant explanation.  To kind a rectangle, we must choose two horizontal sides and two vertical sides. Due to the fact that there space three horizontal lines, we can pick the horizontal political parties in $\binom32$ ways. Since there are 5 vertical lines, us can choose the vertical political parties in $\binom52$ ways. The variety of rectangles we can type is $$\binom32\binom52$$

In general, the variety of rectangles deserve to be developed in a $m \times n$ rectangular grid v $m + 1$ horizontal lines and $n + 1$ vertical lines is the variety of ways us can choose two of the $m + 1$ horizontal lines and also two the the $n + 1$ vertical lines to be the political parties of the rectangle, i m sorry is $$\binomm + 12\binomn + 12 = \frac(m + 1)!(m - 1)!2! \cdot \frac(n + 1)!(n - 1)!2! = \frac(m + 1)m2 \cdot \frac(n + 1)n2$$ ## Not the prize you're feather for? Browse other questions tagged combinatorics discrete-inter-base.netematics permutations combine or questioning your own question.

How countless $6$ number numbers deserve to be created with the number $0, 1, 2$ if the number need to contain at least one $0$? site style / logo © 2021 ridge Exchange Inc; user contributions licensed under cc by-sa. Rev2021.10.1.40358

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