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Math Help - Factorial division problem

  1. #1
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    Factorial division problem

    As a last part of a solution of a second year algebra problem, I have to prove the identity:

    a!b! | (a+b)!

    a factorial times b factorial divides a+b factorial.
    Anyone have any thoughts on how to tackle it? I tried 'writing out' the (a+b)! part, then cutting of either tail of a! or b!, but to no avail....

    Thanks in advance~!

    Tijmen
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  2. #2
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    Hello, TiRune!

    Prove that: . a!b!\,|\,(a+b)!
    I don't have a proof, but I know intuitively that it's true.


    If there are a identical red balls and b identical blue balls,
    . . in how many ways can they be arranged in a row?

    The answer is: . {a+b\choose a,b} \:=\:\frac{(a+b)!}{a!\,b!} . which is an integer.

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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    It suffices to show that every prime power which divides a!b! also divides (a+b)!. Now you can use the fact that the highest power of p which divides n! is p^{\sum_{j=1}^\infty\lfloor \frac{n}{p^j}\rfloor}.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    The product of n consecutive integers is divisible by n!

    Consider the expansion of (a+b)!. It has (a+b) factors.

    The product of the first a factors (or last a factors) is divisible by a!. The product of the last b factors (or first b factors) is divisible by b!.

    Thus, (a+b)! is divisible by a!b!.

    ie a!b! | (a+b)!
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  5. #5
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    Quote Originally Posted by TiRune View Post
    As a last part of a solution of a second year algebra problem, I have to prove the identity:

    a!b! | (a+b)!

    a factorial times b factorial divides a+b factorial.
    Anyone have any thoughts on how to tackle it? I tried 'writing out' the (a+b)! part, then cutting of either tail of a! or b!, but to no avail....

    Thanks in advance~!

    Tijmen
    \frac{(a+b)!}{a!b!}=\frac{(a+b)!}{b!a!}=\frac{(a+b  )!}{([a+b]-a)!a!}=\binom{a+b}{a}=n\ \in\ \mathbb{N}

    as this is "select 'a' elements from 'a+b' elements".
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