How to simplify factorial equations?

\[(2n)! \div n! . n!\]

Post some...

(2n)!÷n!.n! = ??

that doesnt simplify down very nicely, all you can do is: \[\frac{(2n)!}{n!n!} = \frac{n!(n+1)(n+2)\ldots(2n)}{n!n!} = \frac{(n+1)(n+2)\ldots(2n)}{n!}\]

It's choose n from 2n...

right.

Who said u had to simplify it?

Did not get expansion of (2n)! to n!(n+1)(n+2)…(2n)

Reason of simplification is to calculate time complexity of an algorithm whose function calls can be represented in Binary tree!!

If u are in fac, your algorithm ain't gonna be much use....

Its not Fact for sure! :)

(2n n) is bad.....

I know its really bad for sure but its a objective question where complexity is asked.

im using this property to change (2n)! n! = n(n-1)! = n(n-1)(n-2)! = . . .

Way worse than exp...awful.

isn't \[(2n)!=(2n)(2[n-1])(2[n-2]) \cdot \cdot \cdot (2[2])(2[1])\] this the expansion for (2n)!?

its early so i don't know lol

your missing some terms. it would be: \[(2n)! = (2n)(2n-1)(2n-2)(2n-3)\cdots(2)(1)\]

youre*

yep yep you are right

like always joe is always right

O series here: http://www.wolframalpha.com/input/?i=%282n%29!%2F%28n!n!%29

im not always right <.< i just dont like to speak unless i an 95% sure im right lolol If you even seen me in a thread no typing anything, im wrong :P

lol

joe we can do a proof to prove you always correct

<.< lolol

We'll induct him....

Now what's the base case..?

oh, on that note i figured out how to prove: \[(x-y)|x^n-y^n\]by induction, although it still seems retarded and unnecessary since you can do it directly >.>

Gotcha Joe!

nice :)

1 More related question....

OK; so this thread takes care of k and k+1 so assuming he was right about 1+1..QED.

<.<

To determine time complexity, the equation needs to be further simplified to some quadratic, linear (whatever the result may be). What is the simplest way?

I gave u big O series link...

im not going to pretend i know what we are talking about >.>

Think ur supposed to use standard format for step representation, poly, log, quadratic etc...

I don't want to work it all out though...:-) Wolfram can do it better.

Actually I would just say its fac and leave it at that....

Yup!

Thanks Joe and estudies! :) and Best wishes!!

ur welcome.

You could use Stirling's Approximation \( n! \approx \sqrt{2 \pi n} (\frac{n}{e})^{n} \) so \( \frac{(2n)!}{n! n!} \approx (\pi n)^{-\frac{1}{2}} 2^{2n} \) which is O(exponential time)

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