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'

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

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