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Difficult Math Problem #99 - Permutations

This topic has 10 member replies
800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #99 - Permutations

Post Mon Feb 19, 2007 8:15 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
    (A) 38
    (B) 46
    (C) 72
    (D) 86
    (E) 102


    oa coming when some people respond with explanations. from diff math doc

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    Neo2000 Legendary Member
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    Post Mon Feb 19, 2007 8:51 am
    The 5 kids can be arranged among themselves in 5! ways

    Consider the 2 siblings as 1. Then 4 kids can be arranged among themselves in 4! x 2! ways ( since the 2 can be arranged among themselves in 2! ways)

    So Total Number of ways in which 5 kids can be arranged so that 2 particular kids DO NOT sit together = 5! - (4! x 2!)

    Answer Option (C)

    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Mon Feb 19, 2007 1:06 pm
    800guy wrote:
    Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
    (A) 38
    (B) 46
    (C) 72
    (D) 86
    (E) 102


    oa coming when some people respond with explanations. from diff math doc
    Do you mean two couples of siblings or two kids who are siblings ?

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    aim-wsc Legendary Member
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    Post Mon Feb 19, 2007 6:47 pm
    @banona well this is all you have Smile
    work on it.

    by the way you are doing good job at math section.

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    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Feb 19, 2007 6:51 pm
    banona wrote:
    800guy wrote:
    Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
    (A) 38
    (B) 46
    (C) 72
    (D) 86
    (E) 102


    oa coming when some people respond with explanations. from diff math doc
    Do you mean two couples of siblings or two kids who are siblings ?
    this isn't clearly written perhaps, but i was thinking two kids who are siblings.

    Neo2000 Legendary Member
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    Post Mon Feb 19, 2007 6:53 pm
    banona wrote:
    Do you mean two couples of siblings or two kids who are siblings ?
    banona,
    I believe you are misinterpreting the question. The Q only says 2 siblings NOT 2 sets of siblings. You require atleast 2 children to have a sibling pair.

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Feb 19, 2007 6:54 pm
    Neo2000 wrote:
    banona wrote:
    Do you mean two couples of siblings or two kids who are siblings ?
    banona,
    I believe you are misinterpreting the question. The Q only says 2 siblings NOT 2 sets of siblings. You require atleast 2 children to have a sibling pair.
    good eye, neo. i guess the prompt is pretty clear...

    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Tue Feb 20, 2007 2:38 pm
    Neo2000 wrote:
    banona wrote:
    Do you mean two couples of siblings or two kids who are siblings ?
    banona,
    I believe you are misinterpreting the question. The Q only says 2 siblings NOT 2 sets of siblings. You require atleast 2 children to have a sibling pair.
    .

    thank you for the your remark, but you would undestand that I am struggling againt my frenchy-english and subtile stuffs like those usually let me confused,

    thanks again

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    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Tue Feb 20, 2007 3:12 pm
    Answer must be C,
    wHY ?
    let consider the following row ;

    A B C D E
    If five students are to be arranged so that one siblings (2kids) do never sit togoether, so
    the only possibilities are for the siblings to sit on the followings positions ;

    A C
    A D
    A E
    B D
    B E
    C E

    we have at all 6 configurations,
    for each configuation, the two sibilings can switch mutually theirs positions, so that it leads to 12 configurations in sum,

    for each one of these 12 configurations, the other children can sit a number of ways equal to 3! ( WHICH IS THE PEMUTATIONS OF THE
    OTHER CHILDREN AROUND THE VACCANT POSITIONS)

    Then, the total ways are : 12 * 3! = 72

    I believe answer is C.

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    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Tue Feb 20, 2007 3:15 pm
    aim-wsc wrote:
    @banona well this is all you have Smile
    work on it.

    by the way you are doing good job at math section.
    Thank you aim-wsc,
    I hope I will so on the 5th March,
    My God, it is toooooo close !!!!!!!!!

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    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Feb 21, 2007 9:15 am
    oa:

    The total number of ways 5 of them can sit is 120
    when the siblings sit together they can be counted as one entity
    therefore the number of ways that they sit together is 4!=24, but since
    the two siblings can sit in two different ways e.g. AB and BA we multiply 24 by 2 to get the total number of ways in which the 5 children can sit together with the siblings sitting together - 48
    In other words 4P4*2P2
    the rest is obvious 120-48=72

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