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## permutation

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Md.Nazrul Islam Rising GMAT Star
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permutation Sat Mar 31, 2012 5:06 am
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The number of parallelograms that can be formed from a set of four parallel straight line intersecting a set of three parallel straight lines .

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sanju09 GMAT Instructor
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Sat Mar 31, 2012 6:13 am
Md.Nazrul Islam 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 .
I think I tried this in hurry. Please ignore my post and go with Mitch's explanation given below. The correct answer is 18.

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The laws of nature are but the mathematical thoughts of God. ~Euclid

Last edited by sanju09 on Wed Apr 04, 2012 3:03 am; edited 1 time in total

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GMATGuruNY GMAT Instructor
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Sat Mar 31, 2012 6:34 am
Md.Nazrul Islam 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 .
From the set of 4 parallel lines:
We need to choose a combination of 2 to serve as 2 parallel sides of the parallelogram.
The number of combinations of 2 that can formed from 4 choices = 4C2 = 6.

From the set of 3 parallel lines:
We need to choose a combination of 2 to serve as the OTHER 2 parallel sides of the parallelogram.
The number of combinations of 2 that can be formed from 3 choices = 3C2 = 3.

To combine the options above, we multiply:
6*3 = 18.

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ka_t_rin Rising GMAT Star
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Thu Apr 05, 2012 3:30 am
GMATGuruNY wrote:
Md.Nazrul Islam 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 .
From the set of 4 parallel lines:
We need to choose a combination of 2 to serve as 2 parallel sides of the parallelogram.
The number of combinations of 2 that can formed from 4 choices = 4C2 = 6.

From the set of 3 parallel lines:
We need to choose a combination of 2 to serve as the OTHER 2 parallel sides of the parallelogram.
The number of combinations of 2 that can be formed from 3 choices = 3C2 = 3.

To combine the options above, we multiply:
6*3 = 18.
We use combinations and not permutations because we can`t change the order of the lines, right??!

sanju09 GMAT Instructor
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Thu Apr 05, 2012 5:46 am
ka_t_rin wrote:
GMATGuruNY wrote:
Md.Nazrul Islam 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 .
From the set of 4 parallel lines:
We need to choose a combination of 2 to serve as 2 parallel sides of the parallelogram.
The number of combinations of 2 that can formed from 4 choices = 4C2 = 6.

From the set of 3 parallel lines:
We need to choose a combination of 2 to serve as the OTHER 2 parallel sides of the parallelogram.
The number of combinations of 2 that can be formed from 3 choices = 3C2 = 3.

To combine the options above, we multiply:
6*3 = 18.
We use combinations and not permutations because we can`t change the order of the lines, right??!
We use combination because we need to chose.

_________________
The laws of nature are but the mathematical thoughts of God. ~Euclid

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