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Elementary combinatorics

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by Elena89 » Tue Nov 15, 2011 6:21 am
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x- and y-coordinates of P, Q, and R are to be integers that satisfy the inequalities -4 ≤ x ≤ 5 and 6 ≤ y ≤ 16. How many different triangles with these properties could be constructed?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

Kindly solve with reference to the 'Elementary Combinatorics'
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by sl750 » Tue Nov 15, 2011 7:01 am
Number of integers for x coordinate=10
Number of integers for y coordinate=11

The number of coordinates (x,y)for point P can be selected in 10*11
The number of coordinates (x,y) for point Q = 9*11 ; Only value of x changes, as PQ is parallel to the x-axis
The number of coordinates (x,y)for point R = 10*10 ; Only value of y changes
Therefore total number of triangles that can be formed = 10*9*11*10 = 9900
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by Neo Anderson » Tue Nov 15, 2011 8:04 am
Image
refer figure above; the intersections of the lines show the feasible values for x and y.
X coordinate (red color in figure)of P & Q should be the same ; as the triangle is perpendicular at P
Y coordinate (green color in figure)of P & R should be the same ; as PR is parallel to x axis

Thus x coordinate of P can be choosen in 10 ways (as 10 points are therebetween -4 and 5)
y coordinate of P can be choosen in 11 ways (as 11 points are there between 6 and 16)

x coordinate of Q can be choosen in 1 way (as it will be same as x of P)
y coordinate of Q can be choosen in 10 ways (one less because out of 11 one is choosen for P
and Q and P can not have same y coordinate)

x coordinate of R can be choosen in 9 ways (because out of 10 one is choosen for P & Q)
y coordinate of R can be choosen in 1 way (as it will be same as y of P)

Hence 10*11*1*10*9*1=9900 ways

Thus C
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