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 coordinates of P,Q and R are to be integers that satisfy the following inequalities
i. x greater than equal to 4 & less than equal to 5
ii. y is greater than equal to 6 & less than equal to 16.
How many different triangles can be constructed with these properties?
Ans9,900
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Since x and y should be integers,
x: 5  (4) + 1 = 10
y: 16  6 + 1 = 11
we have 10 options for x and 11 options for y. The total number of ways that we can put p on xy plane according to the given ranges for x and y is:
10 Ã— 11 = 110
Since p and r have the same y, once we put p on xy plane, we are no longer concerned about the y of r, we know that yp = yr. So we should only choose x for the point r. Since we selected one x for point p before, now we have one less options for x. In other words, 10  1 = 9.
And point q should have the same x as p. So we have 11  1 = 10 options.
=> the total triangles = 110 Ã— 9 Ã— 10 = 9900
x: 5  (4) + 1 = 10
y: 16  6 + 1 = 11
we have 10 options for x and 11 options for y. The total number of ways that we can put p on xy plane according to the given ranges for x and y is:
10 Ã— 11 = 110
Since p and r have the same y, once we put p on xy plane, we are no longer concerned about the y of r, we know that yp = yr. So we should only choose x for the point r. Since we selected one x for point p before, now we have one less options for x. In other words, 10  1 = 9.
And point q should have the same x as p. So we have 11  1 = 10 options.
=> the total triangles = 110 Ã— 9 Ã— 10 = 9900
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The question should read:
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 yaxes.
The first point we select (in stage 1) dictates the ycoordinate of point R.
In how many ways can we select the xcoordinate of point R?
Well, we can choose any of the 10 coordinates from 4 to 5 inclusive EXCEPT for the xcoordinate 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 yaxes.
The first point we select (in stage 1) dictates the xcoordinate of point Q.
In how many ways can we select the ycoordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the ycoordinate 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) = [spoiler]9900 = C[/spoiler]

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmatcounting?id=775
Then you can try solving the following questions:
EASY
 https://www.beatthegmat.com/whatshould ... 67256.html
 https://www.beatthegmat.com/countingpro ... 44302.html
 https://www.beatthegmat.com/pickinga5 ... 73110.html
 https://www.beatthegmat.com/permutation ... 57412.html
 https://www.beatthegmat.com/simpleonet270061.html
 https://www.beatthegmat.com/mousepelletst274303.html
MEDIUM
 https://www.beatthegmat.com/combinatoric ... 73194.html
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 https://www.beatthegmat.com/digitsnumberst270127.html
 https://www.beatthegmat.com/doubtonsep ... 71047.html
 https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
 https://www.beatthegmat.com/wonderfulp ... 71001.html
 https://www.beatthegmat.com/pscountingt273659.html
 https://www.beatthegmat.com/permutation ... 73915.html
 https://www.beatthegmat.com/pleasesolve ... 71499.html
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Cheers,
Brent
Take the task of building triangles and break it into stages.Right triangle PQR is to be constructed in the xyplane so that the right angle is at P and PR is parallel to the xaxis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities 4 â‰¤ x â‰¤ 5 and 6 â‰¤ y â‰¤ 16. How many different triangles with these properties can be constructed?
(A) 110
(B) 1100
(C) 9900
(D) 10000
(E) 12100
C
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 yaxes.
The first point we select (in stage 1) dictates the ycoordinate of point R.
In how many ways can we select the xcoordinate of point R?
Well, we can choose any of the 10 coordinates from 4 to 5 inclusive EXCEPT for the xcoordinate 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 yaxes.
The first point we select (in stage 1) dictates the xcoordinate of point Q.
In how many ways can we select the ycoordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the ycoordinate 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) = [spoiler]9900 = C[/spoiler]

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmatcounting?id=775
Then you can try solving the following questions:
EASY
 https://www.beatthegmat.com/whatshould ... 67256.html
 https://www.beatthegmat.com/countingpro ... 44302.html
 https://www.beatthegmat.com/pickinga5 ... 73110.html
 https://www.beatthegmat.com/permutation ... 57412.html
 https://www.beatthegmat.com/simpleonet270061.html
 https://www.beatthegmat.com/mousepelletst274303.html
MEDIUM
 https://www.beatthegmat.com/combinatoric ... 73194.html
 https://www.beatthegmat.com/arabianhors ... 50703.html
 https://www.beatthegmat.com/subsetspro ... 73337.html
 https://www.beatthegmat.com/combinatoric ... 73180.html
 https://www.beatthegmat.com/digitsnumberst270127.html
 https://www.beatthegmat.com/doubtonsep ... 71047.html
 https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
 https://www.beatthegmat.com/wonderfulp ... 71001.html
 https://www.beatthegmat.com/pscountingt273659.html
 https://www.beatthegmat.com/permutation ... 73915.html
 https://www.beatthegmat.com/pleasesolve ... 71499.html
 https://www.beatthegmat.com/notwoladie ... 75661.html
 https://www.beatthegmat.com/lanierasco ... 15764.html
Cheers,
Brent

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 Posts: 518
 Joined: Tue May 12, 2015 8:25 pm
 Thanked: 10 times
simply awesome question  seriously.
P = 11*10 ways
R = 9 ways of selecting x
Q = 10 ways of selecting y
11*10*9*10
9900
P = 11*10 ways
R = 9 ways of selecting x
Q = 10 ways of selecting y
11*10*9*10
9900