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

1. anonymous

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

at least for n > 1 ;-)

3. anonymous

Yes, I sort of assumed.

4. anonymous

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

5. nowhereman

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

6. anonymous

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

7. anonymous

thats not right at all, sorry.

8. anonymous

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

9. anonymous

$(n+1)!=(n+1)n!$

10. nowhereman

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

11. anonymous

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

12. anonymous

Fan us, then, bugaloo! ;)