Question

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

Answer 1

No, because, if they were, dividing them would yield 1.

Here,
\[\frac{(2n)!}{2n(n!)}=\frac{2n(2n-1)(2n-2)...(n+1)n!}{2n \times n!}\]\[=(2n-1)(2n-2)...(n+1)\neq1\]

Answer 2

at least for n > 1 ;-)

Answer 3

Yes, I sort of assumed.

Answer 4

how does the (2n)! turn into the (n+1)n! at the end there?

Answer 5

thats because he reduces n! and the first factor that remains is (n+1)n!

Answer 6

does that mean (2n)! = 2n(n+1)n! ?

Answer 7

thats not right at all, sorry.

Answer 8

i am trying to make it look like n! = (n+1)n!, you know?

Answer 9

\[(n+1)!=(n+1)n!\]

Answer 10

yes indeed \[(2n)! = 2n(2n-1)(2n-2)\cdots(2n-(n-1))n!\]

Answer 11

oh i see how that happened now. thanks for spelling it out for me :)

Answer 12

Fan us, then, bugaloo! ;)