Problem Solving

Compare these two questions for a moment:

2. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?


(A) 72
(B) 76
(C) 78
(D) 80
(E) 84

Answer: b

1. Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to x axis. The x and y coordinated of P,Q & R are to be integers that satisfy the inequalities -4<=x<=5 & 6<=y<=16.
How many different triangles wth these properties are possible?

A. 110
B. 1100
C. 9900
D. 10000
E. 12100

Answer: C

Now my question is, can we use OG's method to solve the first problem on the second one? I know how to solve the second problem using combinatorics, which is the most straightforward and efficient approach. But, I am having a bit of difficulty trying to solve it using OG's way from the first problem. Or, check out Mitch's way at the link:

https://www.beatthegmat.com/how-many-tri ... 13159.html





Solving the second problem using OG's method from the first problem:
A triangle must have 3 points; the first point can have a total of 9 choices. (3x3. Since each point must have an integer value, we can have a total of 9 dots on a xy plane.) The second point can have a total of 8 possibilities, since it cannot be the exact same point as the first one. So far, we have 9x8. So for the third point, we should have 7 possibilities. Therefore the answer should be 9 x 8 x 7 -8= 496. (We subtract 8, since the 3 points cannot be in a line.) Of course, this is wrong since 496 > 76. So why is (9x8x7)-8 wrong? Is there a correct way to solve it using OG's way on the second problem?

Thanks!

A triangle must have 3 points; the first point can have a total of 9 choices. (3x3. Since each point must have an integer value, we can have a total of 9 dots on a xy plane.) The second point can have a total of 8 possibilities, since it cannot be the exact same point as the first one. So far, we have 9x8. So for the third point, we should have 7 possibilities. Therefore the answer should be 9 x 8 x 7 -8= 496. (We subtract 8, since the 3 points cannot be in a line.) Of course, this is wrong since 496 > 76. So why is (9x8x7)-8 wrong? Is there a correct way to solve it using OG's way on the second problem?

9*8*7 is wrong. Suppose first you chose (1,1) and then (1,2) and then (2,1) and form a triangle ABC.
Now in next selection suppose you chose (1,2) first, then (2,1) and then (1,1). The triangle formed will be the same. Hence repetition is there.
You can solve it like this:
Total 9 points, to form a triangle select 3. (Order doesn't matter)
Hence 9C3 = 84
Now subtract 8 (3 points in a lint)
Hence 84-8 = 76.

Like cans said, order does not matter in question #2.

Whether you pick (1,1), (1,2) and (2,3) or (2,3), (1,1) and (1,2), you have the same triangle. Since there are 6 ways to pick each triangle (3!), you have divide 9*8*7 by 6. Then you subtract your 8 straight line combinations.

Why is this different from the first question? Because there are constraints on that triangle: it is a right triangle, and one of the sides adjacent to that right angle has to be parallel to the the x-axis. This means that order DOES matter, because the coordinates of the 2nd point depend on the coordinates of the 1st point, and the coordinates of the 3rd point depend on the coordinates of the 2nd and 1st point.

Thank you everyone! I should have taken consideration of the coordinates for each dot on the xy-plane. Thanks for pointing out the overlap, this makes sense now.

And one suggestion..
While posting 2 questions, its better if you mark them in order.
It got a bit confusing considering first question was Q no. 2 and second question was actually Q no. 1
thanks