I tired counting all the rectangles I can check out, however that didn"t work. How perform I technique this?
Go action by step.
First Picture: 1 rectangle
2nd Picture: 2 extra rectangles. The little rectangle, which has actually been included and also the big one, which includes the two small rectangles.
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Third picture: The huge rectangle. Then 2 rectangles, which has 2 little linked rectangles. And the little rectangle, which has actually been added
4th picture: Only one small rectangle.
Fifth Picture: The rectangle, which has the 2 little rectangle and the small extra rectangle.
You go on prefer this. Then amount the amount of rectangles.
answered Jul 20 "14 at 15:34
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Each rectangle has two vertical lines and also 2 horizontal lines.
Tbelow are five vertical lines in the photo, we can label them 1, 2, 3, 4, 5.
If the lefta lot of edge is 1: Then the top and also bottom are uniquely established, and it is basic to check out that 3 or 4 must be the right edge. 2 options.
If the leftthe majority of edge is 2: Then the rightmost edge is 3 or 4 (2 choices), and in either instance tright here are 3 horizontal segments that can serve as the top/bottom($inom32 =3$ choices). So this offers $2 cdot 3 = 6$. 6 options.
If the lefta lot of edge is 3: If the rightthe majority of edge is 5 there is only one rectangle. If the rightthe majority of edge is 4, tbelow are 5 horizontal segments for peak and bottom, so $inom52 = 10$ choices. Hence 11 options.
If the leftmost edge is 4: Then the rightmany edge is 5, and also tright here are four horizontal segments yielding $inom42 = 6$ possible rectanges. 6 options
The total is 2 + 6 + 11 + 6 = 25.
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answered Apr 8 "16 at 22:41
Alex ZornAlex Zorn
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