find the total number of rectangles in a 8x9 chequered board.

- anonymous

find the total number of rectangles in a 8x9 chequered board.

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- schrodinger

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- amistre64

my guess is 72

- amistre64

or do you mean all permutations?

- anonymous

It's not 72.

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## More answers

- anonymous

its not 72. the number is much bigger. this question is under the topic counting/progression. In order to do this you need to find the sequence; i.e. what is no. of rectangles. 1x2, 2x3, 3,4 board.

- amistre64

i figured it wasnt that easy after i posted :) yeah, this is one of those...count the bigger versions of it... type issue

- anonymous

help please! ><

- anonymous

It's a pretty common problem, and am not in the mood to work through it myself, but this gives a pretty good explanation: http://mathforum.org/library/drmath/view/56760.html

- amistre64

its aolt of counting :)

- anonymous

didnt really help.. :/

- anonymous

9C2 x 10C2 isn't a lot of counting.

- anonymous

It's a lot of counting if you apply a medieval method, sure.

- anonymous

anyway. i have another question

- anonymous

find the total number of rectangles in 8x8 board. the answer in the link you gave me is 1296. but i got 196 instead. am i wrong?

- anonymous

I think we can assume you are wrong, yes; you can post an outline of your method for assessment, if you want.

- anonymous

i can explain how i derive mine. by doing the board from 1x1, 2x2,3x3, 4x4, it produces the answer

- anonymous

1x1 -> 0

- anonymous

2x2-> 4
3x3-> 16
4x4-> 36

- anonymous

Firstly, a square is a type of rectangle.

- anonymous

which follows a pattern of \[(2n-2)^{2}\]

- anonymous

my question stats that: think of the best way of specifying a rectangle. therefore i can decide whether to include squares as rectangles.

- anonymous

states*

- anonymous

for my benefit in this case i chose not to.

- anonymous

So it's actually:
1x1 -> 1
2x2 -> 9 (yours, plus the whole thing for 1, plus the 4 1x1s)
3x3 -> 37
The pattern is actually
\[1^3 + 2^3 + 3^3 + ... + ... n^3 \]

- anonymous

Firstly, you cannot decide a square isn't a rectangle. And second, your answer is still wrong.

- anonymous

why is a square a rectangle?

- amistre64

all squares are rectangles; but not all rectangles are squares.

- amistre64

rectangle = opposites sides are the same; and 4, 90 degree corners

- anonymous

There are 1296 rectangles.
There are 204 squares.
There are 1092 rectangles that are not squares.
---
A rectangle is just a 4 sided shape with 4 right angles.

- anonymous

Just a note - there was a typo above -> 3x3 gives 36 not 37.

- anonymous

actually a rectangle is a shape that has 2 equal sides.

- anonymous

which makes square also a rectangle. ok i had that clarified thanks.

- anonymous

Actually, you don't need to say it has two equal sides (and that is not explicit enough - see: parallelogram). I direct you to my definition.

- anonymous

a parallelogram is also a rectangle

- anonymous

No, a rectangle is a parallelogram. The converse is not true - i.e. a parallelogram is not always a rectangle.
Why are you arguing with me? You obviously have no idea who I am. That, and I'm trying to help you.

- anonymous

ok i'm sorry. i googled that and it told me that a parallelogram is a rectangular. guess it came out wrong.

- anonymous

3*3=36. HOW?

- anonymous

*facepalm*

- anonymous

its 1^3 + 2^3 + 3^3

- anonymous

LOL. and newton who are you anyway? ^^

- anonymous

IF IT IS 3^3 THEN ANSWER IS 27 NOT 36???

- anonymous

3x3, in this context, means the number of rectangles on a 3x3 board.
And I'm the person who has never made a Mathematical mistake in their entire life.

- amistre64

at leat no mistakes that anyone has lived to tell about ;)

- anonymous

LOL. are you a lecturer? >.>

- anonymous

YOU R SAYING IT IS NOT A MISTAKE? FUNNY

- anonymous

how do you remove see out of this conversation?

- anonymous

btw newton. can you help me in 8x9 rectangles?

- anonymous

ANY THING WRONG HOLYX?????????????

- anonymous

No, I'm not a lecturer (or anything similar).
And the number of rectangles on an 8 by 8 board can be obtained thus:
There are 10 lines in one direction, and 9 lines in the others. To make a rectangle you need to pick 2 of each, so this is:
\[^10C_2 \times ^9C_2 \text{ where } ^nC_r = \frac{n!}{r!(n-r)!} \]
You can do it by slower method of adding up systematically, but I will not partake in them.

- anonymous

8 by 9*

- anonymous

Ugh, LaTeX fail:
\[^{10}C_2 \]**

- anonymous

you mean 8!/(8-r)!r!

- anonymous

wait im thinking. lol

- anonymous

i dont get it. why 10c2. x 9c2. why do you need to multiply.

- anonymous

Because there are x ways to choose the horizontal lines, and y ways to chose the vertical ones, and therefore x * y ways to combine these.
Say you choose the first set of line horizontally. You can pick any of the sets vertically to go with this. so this would be y ways.
You then move onto the next way to pick vertical lines, and can do each of these with the horizontal ones, so another y ways.
.
You finally get to the xth set of vertical lines, and another y.
Total = x * y

- anonymous

Ugh, I mixed up vertical and horizontal in each part, but hopefully it made sense.

- anonymous

ok yea. 9c2 x 9c2 gives the same aswer 1296!

- anonymous

:)

- anonymous

omg. you're damn smart. how did you thought of that. -,- wtf.

- anonymous

anyway thanks for answering my question (:

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