I exhausted counting every the rectangles I might see, however that didn"t work. Just how do I approach this?

Go action by step.

**First Picture**: 1 rectangle

**Second Picture**: 2 added rectangles. The small rectangle, which has actually been included and the big one, which includes the two little rectangles.

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**Third picture**: The huge rectangle. Then 2 rectangles, which contains 2 small linked rectangles. And also the little rectangle, which has actually been added

**Fourth picture**: only one little rectangle.

**Fifth Picture**: The rectangle, which has the two little rectangle and also the little additional rectangle.

You go on choose this. Then amount the lot of rectangles.

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answered Jul 20 "14 in ~ 15:34

callculus42callculus42

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Each rectangle has actually two upright lines and also two horizontal lines.

There are 5 vertical currently in the picture, we can label lock 1, 2, 3, 4, 5.

**If the leftmost edge is 1:** climate the top and also bottom room uniquely determined, and it is basic to view that 3 or 4 have to be the ideal edge. **2 options**.

**If the outward edge is 2:** climate the rightmost leaf is 3 or 4 (2 choices), and also in either situation there space 3 horizontal segment that can serve together the top/bottom($inom32 =3$ choices). So this offers $2 cdot 3 = 6$. **6 options**.

**If the leftmost edge is 3:** If the rightmost leaf is 5 there is only one rectangle. If the rightmost edge is 4, there space 5 horizontal segments because that top and also bottom, for this reason $inom52 = 10$ choices. Therefore **11 options**.

**If the leftmost edge is 4:** climate the rightmost edge is 5, and there are 4 horizontal segments yielding $inom42 = 6$ feasible rectanges. **6 options**

The full is 2 + 6 + 11 + 6 = 25.

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answered Apr 8 "16 at 22:41

Alex ZornAlex Zorn

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