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How many triangles on the coordinate plane

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by way2ashish » Fri Sep 14, 2012 11:20 pm
Answer is 9C3 - 8 = 84-8 = 76
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by mannvaish » Sun Oct 21, 2012 5:10 am
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.
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by Brent@GMATPrepNow » Sun Oct 21, 2012 5:47 am
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.
Perfect!

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by Brent@GMATPrepNow » Sun Oct 21, 2012 5:51 am
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,
The question asks us to find the total number of triangles.
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

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by kris_na » Mon Nov 26, 2012 5:43 am
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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by oana_m » Tue Jan 08, 2013 10:55 am
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.
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by Brent@GMATPrepNow » Tue Jan 08, 2013 11:03 am
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.
Yep, that's the official answer.

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by bnpetteway » Tue Jan 08, 2013 4:21 pm
@Brent, where did the 6! come from when doing the combinations of it?
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by Brent@GMATPrepNow » Tue Jan 08, 2013 4:30 pm
bnpetteway wrote:@Brent, where did the 6! come from when doing the combinations of it?
First a recap: We have 9 points and we want to select 3 of them (to create triangles).
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

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by Brent@GMATPrepNow » Tue Jan 08, 2013 4:31 pm
Brent@GMATPrepNow wrote:
bnpetteway wrote:@Brent, where did the 6! come from when doing the combinations of it?
First a recap: We have 9 points and we want to select 3 of them (to create triangles).
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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by rajeshsinghgmat » Wed Jan 09, 2013 12:54 am
78

2{2(3*2*2)+(4+4+4+3)}=2(24+15)

=2(39)=78
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by kozmo1 » Tue Jan 29, 2013 6:44 am
Would this be considered a "typical" GMAT question, or a difficult one?
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by Brent@GMATPrepNow » Tue Jan 29, 2013 6:57 am
I'd say it's a difficult one.
700+ level

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by monir.sikder » Thu Jan 31, 2013 9:36 am
very tricky!!! couldn't figure it
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by confusedSoul730 » Wed May 08, 2013 1:51 am
Brent@GMATPrepNow wrote:
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
The first part of your solution looks good (9C3), but we need to subtract more than 4 3-point sets.

Here's my full solution:

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the solution is not visible to me. can you please post it again
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