How many triangles on the coordinate plane
- way2ashish
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The answer is 76.
the logic is thing from all these nine coordinates, when we group 3 points at a time we get 9C3 i.e. 84.
But, in these 84 set of combinations we have some straight kines too, after subtracting those straight line we get our ans.
3 parallel to x-axis.
3 parallel to y-axis.
and 2 diagonal of the figure formed.
So 84-8=76.
Please correct me if I am wrong.
the logic is thing from all these nine coordinates, when we group 3 points at a time we get 9C3 i.e. 84.
But, in these 84 set of combinations we have some straight kines too, after subtracting those straight line we get our ans.
3 parallel to x-axis.
3 parallel to y-axis.
and 2 diagonal of the figure formed.
So 84-8=76.
Please correct me if I am wrong.
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Perfect!mannvaish wrote:The answer is 76.
the logic is thing from all these nine coordinates, when we group 3 points at a time we get 9C3 i.e. 84.
But, in these 84 set of combinations we have some straight kines too, after subtracting those straight line we get our ans.
3 parallel to x-axis.
3 parallel to y-axis.
and 2 diagonal of the figure formed.
So 84-8=76.
Please correct me if I am wrong.
Cheers,
Brent
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The question asks us to find the total number of triangles.jmckenna14 wrote:Why did Brent subtract the number of straight lines from the combination?
I understand there are 84 total combinations because of 9 total points and 3 points to a triangle (subtracting redundancies), so 9!/3! = 84. But why do we subtract the number of straight lines? Confused.
@Brent can you explain?
Thanks,
84 represents the total number of ways to select any 3 points from the 9 given points. However, some of these 3-point selections don't create triangles; they create straight lines.
As such, we need to subtract from 84, all 3-points selections that result in a line
Cheers,
Brent
Hillel wrote:I got (B) 76 :
Actually, there are 9 available points/vertices.
Every triangle needs 3 vertices.
so choose 3 out of 9, this is:
9!
-----
(9-3)!3!
which gives 84.
But we have not to take into account all vertices on the same line
so we have 3 lines of x, 3 lines of y + 2 diagnals.
84-8 = 76
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Yep, that's the official answer.oana_m wrote:IMO: 76
There are 9 points so 9C3 = 84
84-8(3 vertical lines, 3 horizontal, 2 diagonals)= 76
I'd like to know the official answer.
Cheers,
Brent
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First a recap: We have 9 points and we want to select 3 of them (to create triangles).bnpetteway wrote:@Brent, where did the 6! come from when doing the combinations of it?
This can be accomplished in 9C3 ways.
One way to calculate this is to apply the formula: nCr = n!/(r!)(n-r)!
So, 9C3 = 9!/(3!)(9-3)!
= 9!/(3!)(6!)
= 84
Then we continue with the solution (as seen on page 1)..... to get a final answer of 76
If anyone is interested, we have a free video on calculating combinations (like 9C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Cheers,
Brent
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Brent@GMATPrepNow wrote:First a recap: We have 9 points and we want to select 3 of them (to create triangles).bnpetteway wrote:@Brent, where did the 6! come from when doing the combinations of it?
This can be accomplished in 9C3 ways.
One way to calculate this is to apply the combinations formula: nCr = n!/(r!)(n-r)!
So, 9C3 = 9!/(3!)(9-3)!
= 9!/(3!)(6!)
= 84
Then we continue with the solution (as seen on page 1)..... to get a final answer of 76
If anyone is interested, we have a free video on calculating combinations (like 9C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Cheers,
Brent
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the solution is not visible to me. can you please post it againBrent@GMATPrepNow wrote:The first part of your solution looks good (9C3), but we need to subtract more than 4 3-point sets.awesomeusername wrote:A bit tricky.
There are 9 points in the restricted plane. There are three points to a triangle.
9C3 = 9!/3!*6! = 7*8*9/6 = 84
There are four 3 point sets that don't create triangles (when x is the same for all points, or y is the same for all points).
So 84-4 = 80
Here's my full solution: