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If 4 points are indicated on a line and 5 points are indicated on 16 May 2012, 06:52
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D
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75% (01:36) correct 25% (01:52) wrong based on 1774 sessions

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If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90
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D
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Bunuel
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If 4 points are indicated on a line and 5 points are indicated on 16 May 2012, 07:58
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alexpavlos
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!
Show more

Approach #1:

There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\);
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);

Total: \(30+40=70\).

Answer: D.


Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
\(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.
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If 4 points are indicated on a line and 5 points are indicated on 16 May 2012, 07:04
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say line A with 4 points and line B with 5 points. and triangle is PQR.
If P is on line A and QR on line B we can have 4C1*5C2=40 triangles.
If P is on line B and QR on line A we can have 4C2*5C1=30 triangles.
Total =70 triangles.
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If 4 points are indicated on a line and 5 points are indicated on 29 Dec 2012, 01:27
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Not that fun this question :( I wonder how often I would see such a question on test day.

Solution:

* * * * * (line B)

* * * * (line A)

Choose 2 points from line A and a point from line B: = 4!/2!2! * 5!/1!4! = 30
Choose 1 point from line A and 2 points from line B: = 4!/1!3! * 5!/2!3! = 4 * 10 = 40

Combine possiblities: 30 + 40 = 70
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If 4 points are indicated on a line and 5 points are indicated on 19 Jul 2014, 21:45
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See Image below for what my scratch paper looked like on this one
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If 4 points are indicated on a line and 5 points are indicated on 19 Jul 2014, 22:47
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Bunuel
alexpavlos
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\);
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);

Total: \(30+40=70\).

Answer: D.

Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
\(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.
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I really like your 2nd approach :-D
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If 4 points are indicated on a line and 5 points are indicated on 06 Aug 2015, 14:54
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Bunuel
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If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\);
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);

Total: \(30+40=70\).

Answer: D.

Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
\(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.
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Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?
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If 4 points are indicated on a line and 5 points are indicated on 06 Aug 2015, 15:29
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dukenukem
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alexpavlos
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90

Tried doing a few options, but couldn't get it right and the answer provided left me more confused!

Approach #1:
There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\);
With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);

Total: \(30+40=70\).

Answer: D.

Approach #2:

All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
\(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.

Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?
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Combination problems are located at search.php?search_id=tag&tag_id=52
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If 4 points are indicated on a line and 5 points are indicated on 06 Feb 2017, 09:47
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Attached is a visual that should help. Bunuel's method is the most elegant, but here is an alternative strategy if the other solution is not clear.
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If 4 points are indicated on a line and 5 points are indicated on 14 Mar 2018, 16:09
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alex1233
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

A. 20
B. 30
C. 40
D. 70
E. 90
Show more

We have two scenarios:

The first scenario is that the 1 vertex is on the first line and the 2 other vertices are on the second line.
So the number of ways to create such a triangle is:

4C1 x 5C2 = 4 x (5 x 4)/2 = 40

The second scenario is that 2 vertices are on the first line and the 1 vertex is on the second line. So the number of ways to create such a triangle is:

4C2 x 5C1 = (4 x 3)/2 x 5 = 6 x 5 = 30

Thus, the total number of ways to create the triangle is 40 + 30 = 70.

Answer: D
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If 4 points are indicated on a line and 5 points are indicated on 16 Nov 2020, 14:46
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1st line: 4 points

2nd line: 5 points

If we choose 1 point on the first line (4C1) and two points from the second line (5C2), we have 4C1 * 5C2 = 40 triangles.

Now, we can choose 2 points on the first line (4C2) and 1 point on the second line (5C1) to get 4C2 * 5C1 = 6 * 5 = 30 triangles

40 + 30 = 70 triangles

Answer is D.
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If 4 points are indicated on a line and 5 points are indicated on 27 Dec 2020, 20:35
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Out of all the Unique Triangles Possible, every single one of them will have 2 Collinear Points on either the:

-Parallel Line with 5 Dots
or
-Parallel Line with 4 Dots

the 3rd Vertex of the Triangle will have to be on the OTHER Parallel Line in order to create a Triangle with positive area


Case 1: We place the Base on the Bottom Parallel Line with 5 Points

(1st)We need to figure out how many Unique Pairs of 2 Points we can pick out of the 5 Points available on this line.

"5 choose 2" = 5!/2!3! = 10 Unique Combinations

AND

(2nd)For each of these 10 unique combinations, there are 4 Points on the Parallel Line on Top which can combine with the given Pair of Points to create a Triangle with positive area. How many ways can we choose 1 for each of the 10 unique combinations:

"4 choose 1" = 4 ways

(10) * (4) = 40 Unique Triangles in which the "Base" is on the Bottom Parallel Line with 5 Dots


Case 2: We place the Base on the Top Parallel Line with 4 Dots.

Same Logic as above, except the Numbers change:

(1st) "4 choose 2" * "5 choose 1" = [4!/2!2!] * 5 = [6] * 5 = 30 Unique Triangles in which the "Base" is on the Top Parallel Line with 4 Dots


40 Triangles from Case 1 + 30 Triangles from Case 2 =

70 Possible Triangles
(D)
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If 4 points are indicated on a line and 5 points are indicated on 14 May 2023, 09:07
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To form a triangle, we need three non-collinear points.

Among the 9 given points, we have 4 points on one line and 5 points on another parallel line.

To form a triangle, we can choose one point from the line with 4 points and two points from the line with 5 points, or vice versa.

The number of triangles formed is the combination of choosing 1 point from the line with 4 points and 2 points from the line with 5 points, plus choosing 2 points from the line with 4 points and 1 point from the line with 5 points.

Combining these two cases, we have:

C(4, 1) * C(5, 2) + C(4, 2) * C(5, 1)
= 4 * (5 * 4 / (2 * 1)) + (4 * 3 / (2 * 1)) * 5
= 4 * 10 + 6 * 5
= 40 + 30
= 70.

Therefore, the total number of triangles that can be formed using the 9 points is 70.

Hence, the correct answer is D. 70.
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If 4 points are indicated on a line and 5 points are indicated on 15 Nov 2024, 04:48
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I have another way of looking at this question.

Let the line with 4 points be A and 5 points be B for clarifying the solution with you.

So, all together there are 9 points. From these 9 points we can choose 3 points in 9C3 = 84 ways.

Now you cannot have those 3 points in the same line so all 3 points from A OR all 3 points from B is not permissible.

Ways to get all 3 points from A = 4C3 = 4 ways
Ways to get all 3 points from B = 5C3 = 10 ways

So all possible triangles = 84 - (10 +4) = 70 ways.

I like this way better as you can solve triangle problem in rectangular grids this way. Adding all possible from each line can be time consuming for rectangles( 4 or 5 lines cases).
JUST find out total ways - collinear ways (3 points in same line) and you are set for these triangle type geometry combinatorics problems
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