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permutation

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Md.Nazrul Islam Rising GMAT Star
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permutation Post Sat Mar 31, 2012 5:06 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post 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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    Last edited by sanju09 on Wed Apr 04, 2012 3:03 am; edited 1 time in total

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    Post 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 Default Avatar
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    Post 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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    Post 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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