## anonymous 5 years ago find the total number of rectangles in a 8x9 chequered board.

1. amistre64

my guess is 72

2. amistre64

or do you mean all permutations?

3. anonymous

It's not 72.

4. 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.

5. amistre64

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

6. anonymous

7. 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

8. amistre64

its aolt of counting :)

9. anonymous

didnt really help.. :/

10. anonymous

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

11. anonymous

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

12. anonymous

anyway. i have another question

13. 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?

14. anonymous

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

15. anonymous

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

16. anonymous

1x1 -> 0

17. anonymous

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

18. anonymous

Firstly, a square is a type of rectangle.

19. anonymous

which follows a pattern of $(2n-2)^{2}$

20. anonymous

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

21. anonymous

states*

22. anonymous

for my benefit in this case i chose not to.

23. 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 $1^3 + 2^3 + 3^3 + ... + ... n^3$

24. anonymous

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

25. anonymous

why is a square a rectangle?

26. amistre64

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

27. amistre64

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

28. 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.

29. anonymous

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

30. anonymous

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

31. anonymous

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

32. 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.

33. anonymous

a parallelogram is also a rectangle

34. 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.

35. anonymous

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

36. anonymous

3*3=36. HOW?

37. anonymous

*facepalm*

38. anonymous

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

39. anonymous

LOL. and newton who are you anyway? ^^

40. anonymous

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

41. 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.

42. amistre64

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

43. anonymous

LOL. are you a lecturer? >.>

44. anonymous

YOU R SAYING IT IS NOT A MISTAKE? FUNNY

45. anonymous

how do you remove see out of this conversation?

46. anonymous

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

47. anonymous

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

48. 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.

49. anonymous

8 by 9*

50. anonymous

Ugh, LaTeX fail: $^{10}C_2$**

51. anonymous

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

52. anonymous

wait im thinking. lol

53. anonymous

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

54. 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

55. anonymous

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

56. anonymous

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

57. anonymous

:)

58. anonymous

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

59. anonymous

anyway thanks for answering my question (: