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16 Sep 2014, 00:33
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If, out of the six points on the xy-plane, no three of them lie on the same straight line, how many distinct triangles can be formed by selecting three of the six points as vertices?
A. 5
B. 10
C. 20
D. 30
E. 60
A. 5
B. 10
C. 20
D. 30
E. 60
16 Sep 2014, 00:33
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Official Solution:
If, out of the six points on the xy-plane, no three of them lie on the same straight line, how many distinct triangles can be formed by selecting three of the six points as vertices?
A. 5
B. 10
C. 20
D. 30
E. 60
Since no three of the six given points are collinear, any combination of three points will form a triangle. The number of different combinations of three points out of the six is given by \(C_6^3 = \frac{6!}{3! * 3!} = \frac{6 * 5 * 4}{3 * 2} = 20\)
Note that this is not a Geometry question. While it uses basic knowledge of lines and figures, it is actually a probability and combinatorics question. There are 8 questions within GMAT Prep Focus Edition that use similar principles. Here is one example.
Answer: C
If, out of the six points on the xy-plane, no three of them lie on the same straight line, how many distinct triangles can be formed by selecting three of the six points as vertices?
A. 5
B. 10
C. 20
D. 30
E. 60
Since no three of the six given points are collinear, any combination of three points will form a triangle. The number of different combinations of three points out of the six is given by \(C_6^3 = \frac{6!}{3! * 3!} = \frac{6 * 5 * 4}{3 * 2} = 20\)
Note that this is not a Geometry question. While it uses basic knowledge of lines and figures, it is actually a probability and combinatorics question. There are 8 questions within GMAT Prep Focus Edition that use similar principles. Here is one example.
Answer: C
22 Mar 2022, 12:15
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Official Solution:
There are 6 points on the xy-plane. Any 3 points of these 6 don't lie on the same line. How many unique triangles can be drawn using these 6 points as vertices?
A. 5
B. 10
C. 20
D. 30
E. 60
One can try drawing a sketch and counting the number of triangles but this way of solving is not quick enough. You have to guess that it's possible to use the combinations formula. The problem boils down to picking 3 points out of 6. So the answer is \(C_6^3=\frac{6!}{3!*3!} = \frac{6*5*4}{3*2} = 20\).
Answer: C
I agree that using the combinations formula is the most efficient way to solve the problem. Like at least one other person who posted, though, I thought there might be some trick that I was missing, so I checked my work (since I was practicing, not taking a quiz). I did not draw anything, and I did not exactly count triangles. But I did organize information based on 6 unique points: call them A, B, C, D, E, and F. Just maintain alphabetical order, and the process is pretty quick.There are 6 points on the xy-plane. Any 3 points of these 6 don't lie on the same line. How many unique triangles can be drawn using these 6 points as vertices?
A. 5
B. 10
C. 20
D. 30
E. 60
One can try drawing a sketch and counting the number of triangles but this way of solving is not quick enough. You have to guess that it's possible to use the combinations formula. The problem boils down to picking 3 points out of 6. So the answer is \(C_6^3=\frac{6!}{3!*3!} = \frac{6*5*4}{3*2} = 20\).
Answer: C
AB | ABC, ABD, ABE, ABF (4)
AC | ACD, ACE, ACF (3)
AD | ADE, ADF (2)
AE | AEF (1)
That gets us to 10 valid triangles. Notice the 4-3-2-1 pattern. This sort of countdown will continue.
BC | BCD, BCE, BCF (3)
BD | BDE, BDF (2)
BE | BEF (1)
We can add another 6 valid triangles and move on to C.
CD | CDE, CDF (2)
CE | CEF (1)
Another 3 valid triangles to add to the tally. All that is left is the final valid combination.
DE | DEF (1)
10 + 6 + 3 + 1 = 20, so there must be 20 valid triangles within the given conditions. Once you pick up on the pattern with A, the rest goes quickly, and the question can still be solved in under 2 minutes. If you cannot remember which formula to use on test day, why not try logic?
Good luck with your studies.
- Andrew
