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anonymous

  • 5 years ago

is (2n)! the same as 2n(n!)?

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  1. anonymous
    • 5 years ago
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    No, because, if they were, dividing them would yield 1. Here, \[\frac{(2n)!}{2n(n!)}=\frac{2n(2n-1)(2n-2)...(n+1)n!}{2n \times n!}\]\[=(2n-1)(2n-2)...(n+1)\neq1\]

  2. nowhereman
    • 5 years ago
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    at least for n > 1 ;-)

  3. anonymous
    • 5 years ago
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    Yes, I sort of assumed.

  4. anonymous
    • 5 years ago
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    how does the (2n)! turn into the (n+1)n! at the end there?

  5. nowhereman
    • 5 years ago
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    thats because he reduces n! and the first factor that remains is (n+1)n!

  6. anonymous
    • 5 years ago
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    does that mean (2n)! = 2n(n+1)n! ?

  7. anonymous
    • 5 years ago
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    thats not right at all, sorry.

  8. anonymous
    • 5 years ago
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    i am trying to make it look like n! = (n+1)n!, you know?

  9. anonymous
    • 5 years ago
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    \[(n+1)!=(n+1)n!\]

  10. nowhereman
    • 5 years ago
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    yes indeed \[(2n)! = 2n(2n-1)(2n-2)\cdots(2n-(n-1))n!\]

  11. anonymous
    • 5 years ago
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    oh i see how that happened now. thanks for spelling it out for me :)

  12. anonymous
    • 5 years ago
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    Fan us, then, bugaloo! ;)

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