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Director  Joined: 29 Nov 2012
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How many triangles can be inscribed in the heptagon pictured  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 66% (01:17) correct 34% (01:27) wrong based on 344 sessions

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Attachment: heptagon.png [ 15.64 KiB | Viewed 18626 times ]
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210

Originally posted by fozzzy on 12 Jun 2013, 22:38.
Last edited by Bunuel on 12 Jun 2013, 23:28, edited 1 time in total.
Edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 60526
Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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7
6
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210

Generally in a plane if there are $$n$$ points of which no three are collinear, then:
1. The number of triangles that can be formed by joining them is $$C^3_n$$.

2. The number of quadrilaterals that can be formed by joining them is $$C^4_n$$.

3. The number of polygons with $$k$$ sides that can be formed by joining them is $$C^k_n$$.

Since no 3 vertices in given heptagon are collinear, then the number of triangles possible is $$C^3_7=35$$.

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Hope it helps.
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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5
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fozzzy wrote:
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

a) 9
b) 21
c) 35
d) 140
e) 210

You have 7 vertices, you have to pick three vertices to form a triangle.

You can pick the first one in 7 ways, the second in 6 and the third in 5.
$$7*6*5$$, then you divide by $$3!$$ because the order does not matter (triangle ABC=CAB for example)

Tot ways $$\frac{7*6*5}{3!}=35$$, or if you want you can solve it with the formula $$7C3=35$$
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210

I guess another way to look at it is that if we are to consider any line under the heptagon, we can see there are 5 other vertices left with which the line can form a traingle. For example for a line composed of AB, there are vertices C,D,E,F,G with which traingles can be constructed.

In total there are 7 possible lines fo the heptagon. Hence total traingles = 7*5 = 35. Hence [C].

Regards,
A
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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so in the same question if we were asked how many quadrilaterals to be formed it would be 7C4?
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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fozzzy wrote:
so in the same question if we were asked how many quadrilaterals to be formed it would be 7C4?

Yes, that's correct.
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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Bunuel,

I don't seem to remember formulas during exam. Can you please point out the flaw in my approach ?

We are required to make triangles inside the heptagon such that three vertices of the triangle are also vertices of the heptagon.
This means we are simply drawing lines to connect the vertices of the heptagon, connecting 2 vertices of the heptagon at a time (that's how we form the triangles as directed in the problem)

That means we have to choose 2 points at a time from 7 points. That is $$C^2_7=21$$.

Bunuel wrote:
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210

Generally in a plane if there are $$n$$ points of which no three are collinear, then:
1. The number of triangles that can be formed by joining them is $$C^3_n$$.

2. The number of quadrilaterals that can be formed by joining them is $$C^4_n$$.

3. The number of polygons with $$k$$ sides that can be formed by joining them is $$C^k_n$$.

Since no 3 vertices in given heptagon are collinear, then the number of triangles possible is $$C^3_7=35$$.

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Hope it helps.
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Posts: 60526
Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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ajay2121988 wrote:
Bunuel,

I don't seem to remember formulas during exam. Can you please point out the flaw in my approach ?

We are required to make triangles inside the heptagon such that three vertices of the triangle are also vertices of the heptagon.
This means we are simply drawing lines to connect the vertices of the heptagon, connecting 2 vertices of the heptagon at a time (that's how we form the triangles as directed in the problem)

That means we have to choose 2 points at a time from 7 points. That is $$C^2_7=21$$.

Bunuel wrote:
fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210

Generally in a plane if there are $$n$$ points of which no three are collinear, then:
1. The number of triangles that can be formed by joining them is $$C^3_n$$.

2. The number of quadrilaterals that can be formed by joining them is $$C^4_n$$.

3. The number of polygons with $$k$$ sides that can be formed by joining them is $$C^k_n$$.

Since no 3 vertices in given heptagon are collinear, then the number of triangles possible is $$C^3_7=35$$.

Similar questions to practice:
http://gmatclub.com/forum/if-4-points-a ... 32677.html
http://gmatclub.com/forum/abcde-is-a-re ... 86284.html
http://gmatclub.com/forum/abcde-is-a-re ... 33328.html
http://gmatclub.com/forum/the-sides-bc- ... 09690.html
http://gmatclub.com/forum/gmat-diagnost ... 79373.html
http://gmatclub.com/forum/m03-71107.html
http://gmatclub.com/forum/how-many-tria ... 98236.html
http://gmatclub.com/forum/right-triangl ... 88958.html
http://gmatclub.com/forum/how-many-circ ... 28149.html

Hope it helps.

Sorry don't follow you. We need three vertices for a triangle not two.
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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It's alright, I was confusing this with another question. All clear now. Thanks.

Bunuel wrote:
ajay2121988 wrote:
Bunuel,

I don't seem to remember formulas during exam. Can you please point out the flaw in my approach ?

We are required to make triangles inside the heptagon such that three vertices of the triangle are also vertices of the heptagon.
This means we are simply drawing lines to connect the vertices of the heptagon, connecting 2 vertices of the heptagon at a time (that's how we form the triangles as directed in the problem)

That means we have to choose 2 points at a time from 7 points. That is $$C^2_7=21$$.

Sorry don't follow you. We need three vertices for a triangle not two.
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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Bunuel wrote:
fozzzy wrote:
so in the same question if we were asked how many quadrilaterals to be formed it would be 7C4?

Yes, that's correct.

and that will also give the same answer as number of triangles.. as 35 ?
7C4
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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things become interesting when specific number of sides of heptagon is asked 2 include
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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210

Since you need 3 non collinear points to make a triangle

7C3 = 35

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Re: How many triangles can be inscribed in the heptagon pictured  [#permalink]

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fozzzy wrote:
Attachment:
heptagon.png
How many triangles can be inscribed in the heptagon pictured, where the three vertices of the triangle are also vertices of the heptagon?

A. 9
B. 21
C. 35
D. 140
E. 210

Since there are 7 vertices on the heptagon and any of its 3 vertices will form a triangle, then the number of possible triangles is 7C3 = 7!/(3! x 4!) = (7 x 6 x 5)/3! = (7 x 6 x 5)/(3 x 2) = 35.

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