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Number of possible ways to draw a rectangle.

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Number of possible ways to draw a rectangle.

by rahulvsd » Sun Oct 23, 2011 7:43 am
Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

A. 396
B.1260
C.1980
D.7920
E.15840

OA:C
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by Anurag@Gurome » Sun Oct 23, 2011 8:03 am
rahulvsd wrote:Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?
Number of possible x-coordinates = (11 - 3) + 1 = 9
Number of possible y-coordinates = (5 - (-5)) + 1 = 11

Any rectangle drawn in the coordinate plane with sides parallel to the x- and y-axes will have vertices with coordinates of the form (a, b), (c, b), (c, d) and (a, d).

Hence, we can conclude that two x-coordinates (a and c here) and two y-coordinates (b and d here) are enough to define such a rectangle.

Hence, number of such rectangles = (Number of ways to select two x-coordinates from 9)*(Number of ways to select two y-coordinates from 11) = (9C2)*(11C2) = (9*8/2)*(11*10/2) = (11*10*9*2) = (22*90) = 1980

The correct answer is C.
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by moonraker » Sun Oct 23, 2011 8:13 am
The total number of x co-ordinates possible are from x=3 to x=11 ==> total 9 values of x
The total number of y co-ordinates possible are from y=-5 to y=5 ==> total 11 values of y

Now we need only 2 x co-ordinates for x and 2 from y to complete a rectangle (because the sides are parallel to x and y axes )

hence we need to choose 2 x and 2 y co-ordinates......... using combinatorics........

number of ways to choose 2 x co-ordinates from 9 points = 9C2 or 9! / ( (9-2)! * 2!) = 9! / (7! * 2!) ........................... a
number of ways to choose 2 y co-ordinates from 11 points = 11C2 or 11! / ((11-2)! * 2!) = 11! / (9! * 2!) .................... b

Multiplying equations a and b above will give us the total ways of choosing a rectangle = [ 9! / (7! * 2!) ] * [ 11! / (9! * 2!)]

= 11! / ( 7! *2! *2!) = 11 x 10 x 9 x 8 /4 = [spoiler]11x10x9x2 = 1980[/spoiler]

Hence ans = C
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