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Thread: Combinatorial Argument

  1. #1
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    Combinatorial Argument

    Can anyone help get me started on this?

    Provide a combinatorial argument to show that if n and k are positive integers with n = 3k, then n!/(3!)^k is an integer.

    The second step states to "generalize" the result of the above.
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  2. #2
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    Hello Tulki
    Quote Originally Posted by Tulki View Post
    Can anyone help get me started on this?

    Provide a combinatorial argument to show that if n and k are positive integers with n = 3k, then n!/(3!)^k is an integer.

    The second step states to "generalize" the result of the above.
    A well-known result states that the number of arrangements of $\displaystyle n$ items, where $\displaystyle n_1$ of the items are identical of the first kind, $\displaystyle n_2$ items are identical of the second kind, ... is $\displaystyle \frac{n!}{n_1!n_2!...}$. See, for example, just here.

    So if we have $\displaystyle n=3k$ items, which can be sorted into $\displaystyle k$ groups of $\displaystyle 3$, each group containing $\displaystyle 3$ identical items, then the number of arrangements of the $\displaystyle n$ items is $\displaystyle \frac{n!}{(3!)^k}$. This number is therefore an integer.

    To generalise this, the same argument applies if $\displaystyle n = mk$, for any positive integer $\displaystyle m$, whereby $\displaystyle \frac{n!}{(m!)^k}$ is therefore also an integer.

    Grandad
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  3. #3
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    Thanks!

    Luckily, we just had a lecture today which covered this concept, so your contribution makes sense.

    Thanks again for your help.
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