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anonymous
 5 years ago
is (2n)! the same as 2n(n!)?
anonymous
 5 years ago
is (2n)! the same as 2n(n!)?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, because, if they were, dividing them would yield 1. Here, \[\frac{(2n)!}{2n(n!)}=\frac{2n(2n1)(2n2)...(n+1)n!}{2n \times n!}\]\[=(2n1)(2n2)...(n+1)\neq1\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0at least for n > 1 ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I sort of assumed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how does the (2n)! turn into the (n+1)n! at the end there?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0thats because he reduces n! and the first factor that remains is (n+1)n!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that mean (2n)! = 2n(n+1)n! ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats not right at all, sorry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am trying to make it look like n! = (n+1)n!, you know?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0yes indeed \[(2n)! = 2n(2n1)(2n2)\cdots(2n(n1))n!\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see how that happened now. thanks for spelling it out for me :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Fan us, then, bugaloo! ;)
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