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Math Help - No. of ways of arrangement

  1. #1
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    No. of ways of arrangement

    3 boys, 2 girls and a puppy sit at a round table. In how many ways can they be arranged if the puppy is to be seated between any 2 boys?

    Case 1: The girls sit together
    No. of ways=(4-1)!(2!)=12

    Case2: The girls are seperated
    No. of ways=(5-1)!=24

    Total no. of ways=12+24=36
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  2. #2
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    Re: No. of ways of arrangement

    Quote Originally Posted by Punch View Post
    3 boys, 2 girls and a puppy sit at a round table. In how many ways can they be arranged if the puppy is to be seated between any 2 boys?

    Case 1: The girls sit together
    No. of ways=(4-1)!(2!)=12

    Case2: The girls are seperated
    No. of ways=(5-1)!=24

    Total no. of ways=12+24=36
    If sits are not numerated then you have simple equality :

    N_o=2 \cdot 3 \cdot 3 !=36
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  3. #3
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    Re: No. of ways of arrangement

    Can i have reasoning of your workings? It looks simple.
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  4. #4
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    Re: No. of ways of arrangement

    Quote Originally Posted by Punch View Post
    Can i have reasoning of your workings? It looks simple.
    Since there are 3 boys and 2 seats the number of arrangements is : \frac {3 !}{(3-2)! \cdot 2!}=3 , if you don't take in consideration who is on the left hand side from the puppy and who is on the right hand side from the puppy .

    So , to get total number of ways you have to double number : \frac {3 !}{(3-2)! \cdot 2!}

    Now,in each arrangement there are 3 more free seats and 1 boy and 2 girls...so you have 3 ! number of ways in which they can sit .

    Finally , to get total number of arrangements you have to multiply 2 \cdot \frac {3 !}{(3-2)! \cdot 2!} by 3 !
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  5. #5
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    Re: No. of ways of arrangement

    Hello, Punch!

    3 boys, 2 girls and a puppy sit at a round table.
    In how many ways can they be arranged
    if the puppy is to be seated between any 2 boys?

    Assume there are six chairs around the table.

    The puppy can sit anywhere (it doesn't matter).
    There are 3 choices for the boy on its left.
    There are 2 choices for the boy on its right.
    The other 3 people (1 boy, 2 girls) can be seated in 3! ways.

    Therefore: . 3\cdot2\cdot3! \,=\,36 ways.

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