deepakrobi wrote:Can some one explain me the solution.. Sorry i don't have the answer choices.

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?

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 = 110x9x10 = 9900

Cheers,
Brent

For more information about the FCP, here's a free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

If anyone is interested, here's another triangle-counting question. Except this time, there's no restriction regarding right triangles.

https://www.beatthegmat.com/how-many-tri ... 28974.html


Cheers,
Brent

deepakrobi wrote:@Brent, thanks for the answer and this similar question. https://www.beatthegmat.com/how-many-tri ... 28974.html. I am still trying to figure out why there is difference in solution approach for this two problem. The problem mentioned by you doesn't have any right angle restriction and you have to subtract the straight lines from all the possible combinations. but in my problem above we specifically trying to figure out possible ways for each vertex of triangle, couldn't we have used the similar approach (i.e. number of ways for each vertex of triangle) even when triangle is not right angle instead of subtracting all the straight lines? i am little confused, can you please clarify?

The 2 questions aren't meant to be solved in the same way. However, we could use the approach for the other question to solve the original question on this thread.

Warning: this could be extremely time-consuming.

Step 1: Select any 3 points (out of the 110 possible points).
This can be accomplished in 110C3 ways (this is notation for combinations).

Step 2: From the 110C3 ways, subtract all of the selections where we have 3 points in a row. This will eliminate 3-point selections that do not create triangles.
Lots of work here.

At this point, we have the total number of triangles (all triangles).

Step 3: From the remaining triangles, eliminate all of the triangles that are not right triangles.
Even more work here. Too much work if you ask me.

So, just because we have two triangle-counting questions doesn't mean they should be solved using the same manner.

The same applies to using the original approach to solve the question on the other thread. In the other question, we are allowing any kind of triangle, so using the right-triangle approach isn't appropriate.

Cheers,
Brent