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Question - why wouldn't it be 12C1 x 11C1, because you're picking one endpoint, then a second endpoint?
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Bunuel
Chembeti
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.

Answer: B.

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: https://www.gmatprepnow.com/module/gmat- ... /video/789

Cheers,
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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

Final Answer:
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