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Math Help - Combinatorial problem

  1. #1
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    Combinatorial problem

    In how many ways can nine 3s and six 5s be placed in a row so that no two 5s are side by side?

    Please, help me to solve it.
    Thanks!
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  2. #2
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    Re: Combinatorial problem

    Place the fives in a row and place 5 threes between them; this is required by the problem. The rest of the threes can be arbitrarily distributed either between some digits or at either end. The number of such distributions is found using the stars and bars theorem.
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  3. #3
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    Re: Combinatorial problem

    Thanks!
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  4. #4
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    Re: Combinatorial problem

    Hello, Romanka!

    In how many ways can nine 3's and six 5's be placed in a row
    so that no two 5's are side by side?

    Place the nine 3's in a row with a space before, after and between them.

    . . \_\:3\:\_\:3\:\_\:3\:\_\:3\:\_\:3\:\_\:3\:\_\:3 \:\_\:3 \:\_ \:3\:\_


    There are 10 spaces.
    Select 6 of them and insert the 5's.

    There are: . _{10}C_6 \:=\:\frac{10!}{6!\,4!} \:=\:210 ways.
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  5. #5
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    Re: Combinatorial problem

    My solution seems incorrect because it counts several arrangements twice. For example, inserting a 3 between 5 and 3 in 5 3 5 gives the same result as inserting it between 3 and 5. Soroban's solution is correct.
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  6. #6
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    Re: Combinatorial problem

    yes, I see. thanks!
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