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Parallelograms

Expert replies
by MBA.Aspirant » Sat Jul 02, 2011 2:26 pm
How many parallelograms are formed if a set of 10 parallel lines intersects another set of 10 parallel lines?

Thanks
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Source: — Problem Solving |

by sunilrawat » Sat Jul 02, 2011 10:18 pm
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by MBA.Aspirant » Sat Jul 02, 2011 11:08 pm
Please post the explanation. I'm not posting here to know the OA
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by winniethepooh » Sun Jul 03, 2011 4:14 am
@Mba Aspirant
https://www.beatthegmat.com/no-of-parall ... 13682.html
here's the answer, Didn't get Ian's explanation though.
If u understand let me know!
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by Frankenstein » Sun Jul 03, 2011 6:10 am
Hi,
For a parallelogram we need 1 pair of parallel lines from one set and another pair from the other set of parallel lines.
So, from 10 parallel lines, 2lines(1pair) can be selected in 10C2 ways.
Similarly, from the other set of parallel lines.

So, total number of parallel liens is 10C2*10C2 = 45*45 = 2025.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by sunilrawat » Sun Jul 03, 2011 7:14 am
MBA.Aspirant wrote:Please post the explanation. I'm not posting here to know the OA
sorry my friend, I did post my explanation in a different post but I got disconnected at that time.
Although my answer was wrong, what I considered was a grid of (10-1) by (10-1) squares.
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by winniethepooh » Sun Jul 03, 2011 8:11 am
@ frankenstine
Can you please explain me this:
In the following link all I want to know is how are 18 parallelograms formed, looking forward for a reply!
https://www.beatthegmat.com/no-of-parall ... 13682.html :?
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by Frankenstein » Sun Jul 03, 2011 8:25 am
winniethepooh wrote:@ frankenstine
Can you please explain me this:
In the following link all I want to know is how are 18 parallelograms formed, looking forward for a reply!
https://www.beatthegmat.com/no-of-parall ... 13682.html :?
Hi,
The last poster on that thread has already explained it. If your question is how to visualize them on the figure, consider parallelograms of all sizes.
Lets say the dimensions of smallest one are 1x1, we need to count the ones with 1x1,1x2,1x3,2x2,2x3. That is how you get 18 parallelograms.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by amit2k9 » Sun Jul 03, 2011 11:23 pm
each line with two points are required here.
10c2 * 10c2 = 25* 81 = 2025.
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