Lets solve this for y:6

on y = 6, you can choose 10 points to place P and 9 points to place R

9x10 = 90

and for the Q, you have 10 options 7<Q<16

so actually on y=6 you can have 90x10=900 different triangles.

Since we have 11 different y points between 6 and 16

900x11= 9900

Choose C

This must be a tough one.....

possibilities that exist on the X axis = 10

possibilities that exist on the Y axis = 11

so, total possibilities for Point P = 110

Now, since PR has to parallel to X axis, therefore the Y co-ordinate for R will have to be the same as P, and X co-ordinates can be any integer in the range provided, except the point that is already taken up by P.

Hence, possibilities for R = 9

Similarly, PQ will be parallel to the Y axis,
hence, possibilities for Q = 10

Now multiple all to get the total number of combinations...

110 X 9 X 10 = 9900.

Another approach


Let us consider the x co-ordinates 1st :
P can take any of the 10 values between -4 & 5(inclusive)
R can take any ofthe remaining 9 values
x co-ordinate of Q will be same as the x co-ordinate of P

so 10*9*1 = 90


Now the y co-ordinates:
P can take any of the 11 values
Q any of the remaining 10 values
y co-ordinate of R is same as P

so 11*10*1=110


Therefore, total no. of triangles that can be formed is 110*90 = 9900