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no of parallelograms??

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Re: no of parallelograms??

by amitdgr » Wed Jul 09, 2008 1:36 am
umaa wrote:The number of parallelograms that can be formed from a set of FOUR parallel straight line intersecting a set of THREE parallel straight lines=?

Options are,

a. 9
b. 12
c. 6
d. 18
Is the OA c.6 ??


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by umaa » Wed Jul 09, 2008 1:48 am
No my friend. I've got the same answer. The answer is 18 (d).
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by amitdgr » Wed Jul 09, 2008 2:16 am
umaa wrote:No my friend. I've got the same answer. The answer is 18 (d).
18 ? oh .. :shock:

how is it 18 ?


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by Ian Stewart » Wed Jul 09, 2008 4:43 am
You're only counting the six smallest parallelograms in the picture (nice picture, by the way!). If, for example, you take all six of the small parallelograms and consider it to be one shape, you get one large parallelogram. And there are many others:

6 parallelograms that measure 1 across, 1 vertical (smallest possible)
3 parallelograms that are 2 across and 1 vertical
4 parallelograms that are 1 across and 2 vertical
2 parallelograms that are 2 across and 2 vertical
2 parallelograms that are 1 across and 3 vertical
1 parallelogram that is 2 across and 3 vertical (all of them together)

Hope the terminology makes sense. That's a total of 18 parallelograms.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Thanks

by umaa » Wed Jul 09, 2008 5:02 am
Thanks Ian. I missed that out. :lol:
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by amitdgr » Wed Jul 09, 2008 5:05 am
Ian Stewart wrote:You're only counting the six smallest parallelograms in the picture (nice picture, by the way!). If, for example, you take all six of the small parallelograms and consider it to be one shape, you get one large parallelogram. And there are many others:

6 parallelograms that measure 1 across, 1 vertical (smallest possible)
3 parallelograms that are 2 across and 1 vertical
4 parallelograms that are 1 across and 2 vertical
2 parallelograms that are 2 across and 2 vertical
2 parallelograms that are 1 across and 3 vertical
1 parallelogram that is 2 across and 3 vertical (all of them together)

Hope the terminology makes sense. That's a total of 18 parallelograms.
Thanks for explaining it Ian :)

Ian Stewart wrote: (nice picture, by the way!).
:lol:


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by cubicle_bound_misfit » Wed Jul 09, 2008 10:18 pm
you need to choose 2 parallel lines along with 2 vertical line.

2 parallel lines can be chosen from 4 parallel line in 4C2 ways .
2 vertical lines can be chosen from 3 vertical line in 3C2 ways

together the condition can be chosen in 4C2 * 3C2 ways ==18.

Hope that helps.

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Cubicle Bound Misfit
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