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GMAT Diagnostic Test Question 41 [#permalink]
 07 Jun 2009, 01:04
GMAT Diagnostic Test Question 41 Field: combinations Difficulty: 700
How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided polygon? A. 35 B. 40 C. 50 D. 65 E. 70
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Re: GMAT Diagnostic Test Question 41 [#permalink]
13 Jun 2009, 21:54
Explanation
Official Answer: ENumber of triangles can be formed from a polygon with 7 sides = C_7^3 = 35. Number of quadrilaterals can be formed from a polygon with 7 sides = C_7^4 = 35. Total Number of ways that triangles and rectangles can be formed from a polygon with 7 sides = C_7^3 + C_7^4 = 35+35 = 70.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
24 Jul 2009, 05:59
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The question asks about "rectangles" and the answer options speak of "quadrilaterals"
With 4 of the 7 points, there wont be any rectangles formed, i.e all internal angles wont be 90 degrees.
Pls clarify
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Re: GMAT Diagnostic Test Question 41 [#permalink]
24 Jul 2009, 07:06
It should read "quadrilaterals". I'll update the PDF. Thank you! +1 shubhashreeb wrote: The question asks about "rectangles" and the answer options speak of "quadrilaterals"
With 4 of the 7 points, there wont be any rectangles formed, i.e all internal angles wont be 90 degrees.
Pls clarify
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Re: GMAT Diagnostic Test Question 41 [#permalink]
15 Sep 2009, 15:41
How do you know that the number of triangles you can make from a 7 sided polygon is 7c3? I guess I'm a bit confused on how these "triangles" are formed.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
21 Sep 2009, 06:52
"Making" triangles boils down to picking 3 different points out of 7 points that the polygon has. This is why it equals C_7^3. matt0586 wrote: How do you know that the number of triangles you can make from a 7 sided polygon is 7c3? I guess I'm a bit confused on how these "triangles" are formed.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
29 Sep 2009, 04:30
dzyubam wrote: "Making" triangles boils down to picking 3 different points out of 7 points that the polygon has. This is why it equals C_7^3. matt0586 wrote: How do you know that the number of triangles you can make from a 7 sided polygon is 7c3? I guess I'm a bit confused on how these "triangles" are formed. Having just read the solution the question makes sense. But when I just did the test, the actual question didn't make sense to me. I didn't understand how you could form a triangle from another polygon. I think the question should be worded such that it implies that the vertices of the polygon are the vertices of the triangle/quadrilateral. Just my 2c... if everyone else understands it then 
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Re: GMAT Diagnostic Test Question 41 [#permalink]
29 Sep 2009, 06:47
I see your point. Would this rewording be OK: How many triangles and quadrilaterals altogether can be formed from the vertices of a 7-sided polygon?
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Re: GMAT Diagnostic Test Question 41 [#permalink]
29 Sep 2009, 07:07
I'm no GMAT question expert... but I think something along the lines of:
"How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided polygon?"
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Re: GMAT Diagnostic Test Question 41 [#permalink]
06 Oct 2009, 03:03
I guess so... 'cos, with the question in current wordings, I see the answer to be 13.. but again.. with an assumption that we can draw one line to join three any three points in the given polygon.. dzyubam wrote: I see your point. Would this rewording be OK:
How many triangles and quadrilaterals altogether can be formed from the vertices of a 7-sided polygon?
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Re: GMAT Diagnostic Test Question 41 [#permalink]
07 Oct 2009, 07:20
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Re: GMAT Diagnostic Test Question 41 [#permalink]
19 Nov 2009, 17:46
for those who hate formulas:
quadrilaterals : (7x6x5x4)/(1x2x3x4)= 35
triangles : (7x6x5)/(1x2x3)= 35
total 35+35= 70
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Re: GMAT Diagnostic Test Question 41 [#permalink]
22 Dec 2009, 20:58
I have a doubt. Isn't "AND" means multiplying and "OR" means adding up?
I ended doing the same calculations except multiplying them instead of adding and never got the correct answer.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
23 Dec 2009, 05:34
You'd be right if the question was a probability question. Here you just have to find the number of triangles that can be drawn under these circumstances, the number of quadrilaterals, and finally add up the two numbers. The question doesn't ask neither of number of ways these figures can be organized in pairs nor does it ask of a any probability value. I hope it helped. paulnihar5 wrote: I have a doubt. Isn't "AND" means multiplying and "OR" means adding up?
I ended doing the same calculations except multiplying them instead of adding and never got the correct answer.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
12 Sep 2010, 16:35
I got a question, when i solved the problem, i literary drew 7 sided polygon, then started drawing triangles, now from first point it forms 5 triangles, so MULTIPLYING 5*7 GIVES ANSWER 35. but aren't we suppose to omit repeating results? because I interpret the question as if i am suppose to find number of UNIQUE triangles that can be drawn and it comes out to be 20 instead, and 13 UNIQUE quadrilaterals. if that wasn't meant here, i still think asking to find NO. OF UNIQUE triangles and quadrilaterals formed will help complicate problem even more.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
12 Sep 2010, 17:37
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NAKAMONIEL wrote: I got a question, when i solved the problem, i literary drew 7 sided polygon, then started drawing triangles, now from first point it forms 5 triangles, so MULTIPLYING 5*7 GIVES ANSWER 35. but aren't we suppose to omit repeating results? because I interpret the question as if i am suppose to find number of UNIQUE triangles that can be drawn and it comes out to be 20 instead, and 13 UNIQUE quadrilaterals. if that wasn't meant here, i still think asking to find NO. OF UNIQUE triangles and quadrilaterals formed will help complicate problem even more. I don't know how you counted 20 and 13, but C^3_7=35 gives 3 unique points and C^4_7=35 gives 4 unique points, which means that 35 uniques triangles and 35 uniques quadrilaterals can be formed --> 35+35=70. But the answer E (70) to be 100% correct, one thing should be changed in the stem: 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. We see that the above formulas are correct if no 3 points are collinear, so we should mention that our 7-sided polygon is not concave. So I'd suggest to change the stem saying: "... using the vertices of a 7-sided regular polygon". Hope it helps.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
09 Oct 2010, 09:05
basically : 3C7 is the same as 4C7 so the result is quicker calculated by: 2 * 3C7 - it saves few second as there is then only one factorization
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Re: GMAT Diagnostic Test Question 41 [#permalink]
30 Sep 2012, 16:50
Bunuel wrote: NAKAMONIEL wrote: C^3_7=35 Guys, I have a question about the nomenclature used for combinations. In the above quoted text, and in the OA, it says the combination to use is C^3_7=35 . . . . but wouldn't this give us a fraction and lead us to the wrong answer? I believe it should be C^7_3=35 since C^n_r= n!/r!(n-r)! This would then actually equal 35. In the OA, the formula would read: C^3_7=35 which is 3!/7!(-4)! which then equals a fraction and the wrong answer. . . Where is my misunderstanding? Any help would be greatly appreciated.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
01 Oct 2012, 08:18
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Krest19 wrote: Bunuel wrote: NAKAMONIEL wrote: C^3_7=35 Guys, I have a question about the nomenclature used for combinations. In the above quoted text, and in the OA, it says the combination to use is C^3_7=35 . . . . but wouldn't this give us a fraction and lead us to the wrong answer? I believe it should be C^7_3=35 since C^n_r= n!/r!(n-r)! This would then actually equal 35. In the OA, the formula would read: C^3_7=35 which is 3!/7!(-4)! which then equals a fraction and the wrong answer. . . Where is my misunderstanding? Any help would be greatly appreciated. and C^7_3 are just different ways of writing the same: choosing 3 out of 7. Hope it's clear.
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Re: GMAT Diagnostic Test Question 41 [#permalink]
12 Jan 2013, 10:02
safari83 wrote: for those who hate formulas:
quadrilaterals : (7x6x5x4)/(1x2x3x4)= 35
triangles : (7x6x5)/(1x2x3)= 35
total 35+35= 70 Yep, this is the safest way for me to understand it; factorial possibilities over chosen ones.
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Re: GMAT Diagnostic Test Question 41
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12 Jan 2013, 10:02
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