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If 4 points are indicated on a line and 5 points are

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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!
[Reveal] Spoiler: OA

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If 4 points are indicated on a line and 5 points are [#permalink]

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New post 16 May 2012, 07:58
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alexpavlos wrote:
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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Re: If 4 points are indicated on a line and 5 points are indicat [#permalink]

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New post 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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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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New post 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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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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See Image below for what my scratch paper looked like on this one
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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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New post 06 Aug 2015, 15:29
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dukenukem wrote:
Bunuel wrote:
alexpavlos wrote:
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?


Combination problems are located at search.php?search_id=tag&tag_id=52

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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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New post 19 Jul 2014, 22:47
Bunuel wrote:
alexpavlos wrote:
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.


I really like your 2nd approach :-D
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Re: If 4 points are indicated on a line and 5 points are [#permalink]

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New post 06 Aug 2015, 14:54
Bunuel wrote:
alexpavlos wrote:
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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If 4 points are indicated on a line and 5 points are [#permalink]

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New post 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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Screen Shot 2017-02-06 at 8.46.05 AM.png [ 117.98 KiB | Viewed 2957 times ]


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