27. How countless squares room there ~ above a chessboard or chequerboard?? (the prize is no 64)Can you expand your method to calculation the number of rectangles on a chessboard?

Another puzzle the was e-mailed to me through this website. Mine instinct was the the answer was just a lot, however I thought around it and the equipment is actually reasonably simple...

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Before reading the answer have the right to I attention you in a clue?The very first thing is why the answer is not simply 64... All the red squares in the above snapshot would count as valid squares, therefore we room asking how many squares of any kind of dimension from 1x1 to 8x8 there room on a chess board.The crucial is to think how countless positions there room that each size of square can be located... A 2x2 square, because that example, can, by virtue of it"s size, be situated in 7 places horizontally and also 7 areas vertically. In other words in 49 different positions. A 7x7 square though have the right to only right in 2 positions vertically and also 2 horizontally. Think about what"s below... sizehorizontal positionsvertical positionspositions204
1x18864
2x27749
3x36636
4x45525
5x54416
6x6339
7x7224
8x8111
total
In total there space 204 squares ~ above a chessboard. This is the sum of the variety of possible positions for all the squares of size 1x1 to 8x8.

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## Formula for n x n Chessboard?

It"s clean from the analysis over that the systems in the situation of n x n is the sum of the squares from n2 to 12 that is come say n2 + (n-1)2 + (n-2)2 ... ... 22 + 12Mathematically that is composed as follows: The evidence of the explicit systems is past the border of this site, but if you desire to watch it up a mathematician would refer to it as "the amount of the squares the the an initial n organic numbers." The last answer is offered byn3/3 + n2/2 + n/6

## Can you prolong your an approach to calculate the number of rectangles ~ above a chessboard?

Below are some examples of feasible rectangles... All the the over examples would be vailid rectanges...There is an ext than one means of solving this. However it makes sense to extend our an approach from the squares difficulty first. The crucial to this is come think of every rectangle individually and consider the variety of positions it have the right to be located. For instance a 3x7 rectangle have the right to be located in 6 location horizontally and 2 vertically. Native this we can develop a procession of all the feasible rectangles and also sum. 1296
 Dimensions 1 2 3 4 5 6 7 8 Positions 8 7 6 5 4 3 2 1 1 8 64 56 48 40 32 24 16 8 2 7 56 49 42 35 28 21 14 7 3 6 48 42 36 30 24 18 12 6 4 5 40 35 30 25 20 15 10 5 5 4 32 28 24 20 16 12 8 4 6 3 24 21 18 15 12 9 6 3 7 2 16 14 12 10 8 6 4 2 8 1 8 7 6 5 4 3 2 1
In total then there space 1296 feasible rectangles.

## Elegant approach to rectangles, consider the vertices and also diagonals. I"ve been sent an imaginative solution come the trouble of the variety of rectangles top top a chessboard through Kalpit Dixit. This solution tackles the worry from a various approach. Rather than looking at particular sizes the rectangles and working out whereby they deserve to be located we start at the other end and look at areas first.The vertices are the intersections. Because that our chessboard there space 81 (9 x 9). A diagonal starting at one vertex and ending at another will uniquely define a rectangle. In stimulate to it is in a diagonal and also not a upright or horizontal heat we might start anywhere but the end point must not have the very same vertical or horizontal coordinate. Thus there are 64 (8 x 8) possible end points.There are as such 81 x 64 = 5184 agree diagonals.However, whilst each diagonal describes a distinctive rectangle, each rectangle go not describe a unique diagonal. We view trivially that each rectangle have the right to be represented by 4 diagonals.So our variety of rectangles is provided by 81 x 64 /4 = 1296

## n x n or n x m?

The n x n (eg. 9x9,) or n x m (eg 10x15,) difficulties can now be calculated. The variety of vertices being offered by (n + 1)2 and (n + 1).(m + 1) respectively. Hence the last solutions room as follows.n x n: (n + 1)2 x n2 / 4n x m: (n + 1) x (m + 1) x (n x m) / 4Which can obviously be arranged right into something an ext complicated.

## Rectangles in Maths Nomenclature

It"s constantly my intention to explain the troubles without formal maths nomenclature, v reasoning and common sense. But there is fairly a practiced solution below if you perform know around combinations, together in permutations and also combinations. Horizontally us are selecting 2 vertices indigenous the 9 available. The order does not issue so it"s combinations rather than permutations. And also the exact same vertically. Therefore the answer to the rectangle difficulty can be answered by:9C2•9C2 = 362 = 1296
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