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D01-40

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D01-40  [#permalink]

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New post 15 Sep 2014, 23:13
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E

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Re D01-40  [#permalink]

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New post 15 Sep 2014, 23:13
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Official Solution:

How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?

A. 35
B. 40
C. 50
D. 65
E. 70

Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of \(C^3_7+C^4_7=35+35=70\) different triangles and quadrilaterals can be formed.

Answer: E
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D01-40  [#permalink]

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New post 01 Jan 2015, 20:03
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I think it is very hard to understand what this question is asking. I never would have been able to understand this premise regardless of time spent on it.
Is the seven sided regular polygon a regular star or a heptagon? You cannot form any triangles or quadrilaterals using only the angle of the heptagon.
Finally, triangles and quadrilaterals aren't made from vertices, but from vertices and sides. There could be infinitely many side lengths, therefore infinitely many triangles and quadrilaterals.
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Re: D01-40  [#permalink]

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New post 31 Oct 2015, 20:09
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mcolbert wrote:
I think it is very hard to understand what this question is asking. I never would have been able to understand this premise regardless of time spent on it.
Is the seven sided regular polygon a regular star or a heptagon? You cannot form any triangles or quadrilaterals using only the angle of the heptagon.
Finally, triangles and quadrilaterals aren't made from vertices, but from vertices and sides. There could be infinitely many side lengths, therefore infinitely many triangles and quadrilaterals.


it is a regular polygon with 7 sides. Regular polygons have equal sides, and equal angles.

for the sake of understanding, draw a 7 sides polygon and start drawing triangles and quadrilaterals.
we clearly see that from one angle, we can draw lines and form 4 triangles. How many triangles can be? well, we have 7 sides, and from 7 sides, we can have 4 triangles
4C7 = 7!/4!3! = 5*6*7/1*2*3 = 35
now, draw quadrilaterals. From each angle, we can draw lines and form 3 quadrilaterals. How many quadrilaterals can be drawn? 7 sides, 3 quadrilaterals...
3C7 = which is equal to 4C7
now we have 35+35 = 70.
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Re D01-40  [#permalink]

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New post 20 Jun 2016, 15:47
I think this is a high-quality question and I agree with explanation.
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Re: D01-40  [#permalink]

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New post 02 Jul 2016, 20:18
mvictor wrote:
mcolbert wrote:
How many triangles can be? well, we have 7 sides, and from 7 sides, we can have 4 triangles

now, draw quadrilaterals. From each angle, we can draw lines and form 3 quadrilaterals. How many quadrilaterals can be drawn? 7 sides, 3 quadrilaterals...


I drew the lines and ended up with 5 triangles (not 4) and 4 quadrilaterals (not 3): Red, Purple, Green, and Blue.
Check the images that I drew.
Please, point out, where is the mistake.

Thank you.
>> !!!

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Re: D01-40  [#permalink]

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New post 03 Jul 2016, 02:45
AndreiGMAT wrote:
mvictor wrote:
mcolbert wrote:
How many triangles can be? well, we have 7 sides, and from 7 sides, we can have 4 triangles

now, draw quadrilaterals. From each angle, we can draw lines and form 3 quadrilaterals. How many quadrilaterals can be drawn? 7 sides, 3 quadrilaterals...


I drew the lines and ended up with 5 triangles (not 4) and 4 quadrilaterals (not 3): Red, Purple, Green, and Blue.
Check the images that I drew.
Please, point out, where is the mistake.

Thank you.


As an example I'll show you 2 triangles that you missed:
Attachment:
Untitled.png
You are missing triangles and quadrilaterals that can be formed from other vertices.
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Re: D01-40  [#permalink]

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New post 03 Jul 2016, 03:15
I think you missing 1 triangle and 1 quadrilateral, Please, see your quote:

Bunuel wrote:
Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of C37+C47=35+35=70C73+C74=35+35=70 different triangles and quadrilaterals can be formed.


In my images I stated that you can form 5 triangles from one angle, and 4 quadrilaterals from one angle. While You formed 4 triangles and 3 quadrilaterals.

So shouldn't be the solution:

7C5+7C4=21+35=56 different triangles and quadrilaterals can be formed
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New post 03 Jul 2016, 03:26
AndreiGMAT wrote:
I think you missing 1 triangle and 1 quadrilateral, Please, see your quote:

Bunuel wrote:
Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of C37+C47=35+35=70C73+C74=35+35=70 different triangles and quadrilaterals can be formed.


In my images I stated that you can form 5 triangles from one angle, and 4 quadrilaterals from one angle. While You formed 4 triangles and 3 quadrilaterals.

So shouldn't be the solution:

7C5+7C4=21+35=56 different triangles and quadrilaterals can be formed


Sorry I cannot draw all 35 triangles and 35 quadrilaterals but the logic in my solution explains why it is so. Any 3 distinct points from 7 will form a triangle and any 4 distinct points from 7 will form a quadrilateral.
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Re: D01-40  [#permalink]

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New post 03 Jul 2016, 05:50
Bunuel wrote:

Sorry I cannot draw all 35 triangles and 35 quadrilaterals but the logic in my solution explains why it is so. Any 3 distinct points from 7 will form a triangle and any 4 distinct points from 7 will form a quadrilateral.


Bunuel, if you could explain this method just algebraically for square and pentagon (how many triangles can be formed), so I could better understand your logic.

Thanks
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New post 03 Jul 2016, 08:36
AndreiGMAT wrote:
Bunuel wrote:

Sorry I cannot draw all 35 triangles and 35 quadrilaterals but the logic in my solution explains why it is so. Any 3 distinct points from 7 will form a triangle and any 4 distinct points from 7 will form a quadrilateral.


Bunuel, if you could explain this method just algebraically for square and pentagon (how many triangles can be formed), so I could better understand your logic.

Thanks


Below are cases for square. We can make only 1 (4C4 = 1) quadrilateral using the vertices of a square and we can make 4 triangles (4C3 = 4).

Attachment:
Untitled.png

>> !!!

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Re: D01-40  [#permalink]

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New post 24 Mar 2017, 09:43
Bunuel wrote:
Official Solution:

How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?

A. 35
B. 40
C. 50
D. 65
E. 70

Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of \(C^3_7+C^4_7=35+35=70\) different triangles and quadrilaterals can be formed.

Answer: E

would the answer be same even if it was not a regular polygon
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Re: D01-40  [#permalink]

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New post 24 Mar 2017, 09:58
KARISHMA315 wrote:
Bunuel wrote:
Official Solution:

How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?

A. 35
B. 40
C. 50
D. 65
E. 70

Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of \(C^3_7+C^4_7=35+35=70\) different triangles and quadrilaterals can be formed.

Answer: E

would the answer be same even if it was not a regular polygon


The fact that the polygon is regular guarantees that no 3 of the vertices are not collinear.
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Re: D01-40  [#permalink]

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New post 15 Jul 2017, 07:00
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Hello Everyone,
What would happen if question would be formulated as below?
How many Distinct triangles and Distinct quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?
What approach we need to keep in mind for solving such question.
Thanks in advance.
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New post 15 Jul 2017, 08:39
For triangle it's 7c3-7 and for quadrilateral it's 7c4-7 . Adding both gives 70. Hence E.

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Re: D01-40  [#permalink]

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New post 22 Apr 2018, 05:42
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Bunuel wrote:
How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?

A. 35
B. 40
C. 50
D. 65
E. 70


TRIANGLES
If you choose any 3 of the 7 vertices, you can connect them with lines to create a unique triangle.
So, the question becomes "In how many different ways can we select 3 vertices from 7 vertices?"
Since the order in which we select the 3 vertices does not matter, we can use COMBINATIONS.
We can select 3 vertices from 7 vertices in 7C3 ways.
7C3 = 35

ASIDE: we have a video on calculating combinations (like 7C3 and 7C4) in your head - see below


QUADRILATERALS
If you choose any 4 of the 7 vertices, you can connect them with lines to create a unique quadrilateral.
So, in how many different ways can we select 4 vertices from 7 vertices?
Since the order in which we select the 4 vertices does not matter, we can use COMBINATIONS.
We can select 4 vertices from 7 vertices in 7C4 ways.
7C4 = 35

So, the TOTAL number of triangles and quadrilaterals possible = 35 + 35 = 70

Answer: E

Cheers,
Brent

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New post 06 Aug 2018, 10:21
Hello?
Thanks for the question and the explanation however i think this is a very tricky question.
Usually in combinatorics (i have read it several times) when asked an AND as for this question (triangles And quadrilaterals), we should multiply the result and when it is an OR we should add.
Why did we add in this case? Why the previous rule doesn't work here?
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New post 10 Aug 2018, 21:06
The question should ask "How many triangles OR (not and) quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?" Please help where my thought process going wrong.
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Re: D01-40  [#permalink]

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New post 28 Oct 2018, 18:33
I think the wording is not correct here. It should be triangles OR quadrilaterals.
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Re: D01-40 &nbs [#permalink] 28 Oct 2018, 18:33
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