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Set S contains nine distinct points in the coordinate plane.

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Set S contains nine distinct points in the coordinate plane.

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Source: Veritas Prep

Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?

A. 56
B. 70
C. 74
D. 79
E. 84

The OA is C.

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BTGmoderatorLU wrote:
Source: Veritas Prep

Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?

A. 56
B. 70
C. 74
D. 79
E. 84
To form a triangle, we must select 3 points such that at most 2 are collinear.

Case 1: Select 3 points not on the x-axis
From the 4 points not on the x-axis, the number of ways to choose 3 = 4C3 = (4*3*2)/(3*2*1) = 4.

Case 2: Select 1 point on the x-axis and two points not on the x-axis
From the 5 points on the x-axis, the number of ways to choose 1 = 5C1 = 5.
From the 4 points not on the x-axis, the number of ways to choose 2 = 4C2 = (4*3)/(2*1) = 6.
To combine these options, we multiply:
5*6 = 30.

Case 3: Select 2 points on the x-axis and 1 point not on the x-axis
From the 5 points on the x-axis, the number of ways to choose 2 = 5C2 = (5*4)/(2*1) = 10.
From the 4 points not on the x-axis, the number of ways to choose 1 = 4C1 = 4.
To combine these options, we multiply:
10*4 = 40.

Total ways = Case 1 + Case 2 + Case 3 = 4 + 30 + 40 = 74.

The correct answer is C.

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BTGmoderatorLU wrote:
Source: Veritas Prep

Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?

A. 56
B. 70
C. 74
D. 79
E. 84
\[? = C\left( {9,3} \right) - C\left( {5,3} \right) = \frac{{9 \cdot 8 \cdot 7}}{{3 \cdot 2}} - \frac{{5 \cdot 4}}{2} = 12 \cdot 7 - 10 = 74\]
This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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BTGmoderatorLU wrote:
Source: Veritas Prep

Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?

A. 56
B. 70
C. 74
D. 79
E. 84
We have 5 (collinear) points that are on the x-axis and 4 points that are not on the x-axis. Since no set of three points in S is collinear except those on the x-axis, we can have the following 3 cases forming a triangle: 1) two points on the x-axis and one point not on the x-axis, 2) one point on the x-axis and two points not on the x-axis, and 3) three points not on the x-axis.

Case 1: Two points on the x-axis and one point not on the x-axis

There are 5C2 x 4C1 = 10 x 4 = 40 such triangles in this case.

Case 2: one point on the x-axis and two points not on the x-axis

There are 5C1 x 4C2 = 5 x 6 = 30 such triangles in this case.

Case 3: Three points not on the x-axis

There are 4C3 = 4 such triangles in this case.

Therefore, there are 40 + 30 + 4 = 74 triangles that can be formed.

Answer: C

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Founder and CEO
scott@targettestprep.com



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