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# How many circles can be drawn on the line segment

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How many circles can be drawn on the line segment [#permalink]

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26 Feb 2012, 00:05
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Question Stats:

75% (01:48) correct 25% (01:12) wrong based on 115 sessions

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Assume that there are 12 points on a straight line as shown below:

--*-*-*----*-*-*-*-*--*-*-*-*

How many circles can be drawn on the line segment above if each circle is subject to the rule that the endpoints of one of its diameters must be any two of the points?

36
66
72
78
121
[Reveal] Spoiler: OA

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-Aravind Chembeti

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Re: How many circles can be drawn on the line segment [#permalink]

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26 Feb 2012, 00:15
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Chembeti wrote:
Assume that there are 12 points on a straight line as shown below:

--*-*-*----*-*-*-*-*--*-*-*-*

How many circles can be drawn on the line segment above if each circle is subject to the rule that the endpoints of one of its diameters must be any two of the points?

A. 36
B. 66
C. 72
D. 78
E. 121

Any 2 different points out of given 12 colinear points can be the endpoints of the diameter, so there can be $$C^2_{12}=66$$ circles drawn.

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Re: How many circles can be drawn on the line segment [#permalink]

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23 Oct 2013, 13:05
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Re: How many circles can be drawn on the line segment [#permalink]

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23 Oct 2013, 19:10
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This is an interesting way to disguise a combination problem. As Bunuel noted, you can create a circle through any two points, so what you really have is a question of how many unique ways to select pairs from a set of 12 points, or:

12!/2!(12-2)! --> 12!/2!10! --> (12*11)/2 --> 6*11 = 66
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Re: How many circles can be drawn on the line segment [#permalink]

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25 Nov 2015, 12:40
Hello from the GMAT Club BumpBot!

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Re: How many circles can be drawn on the line segment [#permalink]

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26 Nov 2015, 13:38
Question - why wouldn't it be 12C1 x 11C1, because you're picking one endpoint, then a second endpoint?
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Re: How many circles can be drawn on the line segment [#permalink]

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26 Nov 2015, 14:22
Bunuel wrote:
Chembeti wrote:
Assume that there are 12 points on a straight line as shown below:

--*-*-*----*-*-*-*-*--*-*-*-*

How many circles can be drawn on the line segment above if each circle is subject to the rule that the endpoints of one of its diameters must be any two of the points?

A. 36
B. 66
C. 72
D. 78
E. 121

Any 2 different points out of given 12 colinear points can be the endpoints of the diameter, so there can be $$C^2_{12}=66$$ circles drawn.

Bunuel's solution is the best/fastest approach I can think of.

If anyone is interested, we have a free video on quickly calculating combinations (like 12C2) in your head: http://www.gmatprepnow.com/module/gmat- ... /video/789

Cheers,
Brent
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Re: How many circles can be drawn on the line segment [#permalink]

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28 Nov 2015, 20:40
Hi All,

There's a great built-in pattern that you can use to solve this question with just a bit of 'brute force' arithmetic.

Let's call the points A,B,C,D,E F,G,H,I,J K and L

The circles that include point A are...
AB, AC, AD, AE, AF, AG, AH, AI, AJ, AK and AL = 11 circles.

Since AB has already been accounted for, here are the additional circles that contain point B...
BC, BD, BE, BF, BG, BH, BI, BJ, BK and BL = 10 circles

Since AC and BC have already been accounted for, here are the additional circles that contain point C...
CD, CE, CF, CG, CH, CI, CJ, CK and CL = 9 circles

Notice how the numbers decrease by 1 with each set of new circles? That pattern will continue, so we have...

11+10+9+8+7+6+5+4+3+2+1 = 66 total circles

[Reveal] Spoiler:
B

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Re: How many circles can be drawn on the line segment   [#permalink] 28 Nov 2015, 20:40
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