TOUGH!!

Take the task of building triangles and break it into stages.

Stage 1: Select any point where the right angle will be (point P).
The point can be selected from a 10x11 grid. So, there 110 points to choose from.
This means that stage 1 can be completed in 110 ways.

Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the y-coordinate of point R.
In how many ways can we select the x-coordinate of point R?
Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1).
So, there are 9 coordinates to choose from.
This means that stage 2 can be completed in 9 ways.

Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the x-coordinate of point Q.
In how many ways can we select the y-coordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1).
So, there are 10 coordinates to choose from.
This means that stage 3 can be completed in 10 ways.

So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = 9900

Answer: C

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x-axis: P and Q will always have same x coordinate so lets not consider Q in this scenario

P : can have values from -4 to 5. So, P can be placed in any one of the 10 coordinates
R : can also have values from -4 to 5 but P has already taken a point and R cannot take the same position because then it wont be a triangle.
So R can be placed in one of the 9 coordinates.

So in x-axis alone you can have 10*9 bases(PR) of the triangle.

y-axis : P and R will always have same x coordinate so lets not consider R in this scenario

P : can have values from 6 to 16. So, P can be placed in any one of the 11 coordinates
Q : can also have values from 6 to 16 but P has already taken a point and Q cannot take the same position because then it wont be a triangle anymore.
So Q can be placed in one of the 10 coordinates.

So in y-axis alone you can have 11*10 bases(PQ) of the triangle.

Thus total triangle you can have is 9*10*11*10 = 9900.

To approach this, we have to find ways to select each of the coordinates (x,y) of Points P, Q, and R.

Such as P, Q, and R follow the x and y constraints

Anyways, first let's find out the total possible points we can select for both x, and y

for x, we have 10 possible points
for y, we have 11 possible points

Now, Let's check the number of ways Q can take values.

X coordinate of Q = 10C1 ways
Y coordinate of Q = 11C1 ways

Q is now dealt with

For P, the x coordinate is now fixed (as PQR is a right angle triangle and QP is perpendicular to PR, which is parallel to the X axis) because QP is parallel to the y-axis

but we still need to decide a y coordinate -

Therefore,
Y coordinate of P = 10C1 ways (One point is occupied by Q)

Now for R. We know, its y coordinate is fixed as PR is parallel to X axis

Therefore, for its Y coordinate =
Y coordinate of R = 9C1 ways (one point is occupied by P)


Therefore, total combinations =
[(X coordinate of Q = 10C1 ways) * (Y coordinate of Q = 11C1 ways)] * [Y coordinate of P = 10C1 ways] * [Y coordinate of R]
= 10*11*10*9
= 9,900

Thank You

Sincerely
Daksh Kumar