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permutation

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permutation

by Md.Nazrul Islam » Sat Mar 31, 2012 4:06 am
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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by sanju09 » Sat Mar 31, 2012 5:13 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 [spoiler]18[/spoiler].
Last edited by sanju09 on Wed Apr 04, 2012 2:03 am, edited 1 time in total.
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by GMATGuruNY » Sat Mar 31, 2012 5:34 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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by ka_t_rin » Thu Apr 05, 2012 2:30 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??!
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by sanju09 » Thu Apr 05, 2012 4:46 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.
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