Question

How to simplify factorial equations?

Answer 1

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

Answer 2

Post some...

Answer 3

(2n)!÷n!.n! = ??

Answer 4

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

Answer 5

It's choose n from 2n...

Answer 6

right.

Answer 7

Who said u had to simplify it?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Its not Fact for sure! :)

Answer 12

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

Answer 13

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

Answer 14

im using this property to change (2n)!

n! = n(n-1)! = n(n-1)(n-2)! = . . .

Answer 15

Way worse than exp...awful.

Answer 16

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

Answer 17

its early so i don't know lol

Answer 18

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

Answer 19

youre*

Answer 20

yep yep you are right

Answer 21

like always joe is always right

Answer 22

O series here:

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

Answer 23

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

Answer 24

lol

Answer 25

joe we can do a proof to prove you always correct

Answer 26

<.< lolol

Answer 27

We'll induct him....

Answer 28

Now what's the base case..?

Answer 29

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 >.>

Answer 30

Gotcha Joe!

Answer 31

nice :)

Answer 32

1 More related question....

Answer 33

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

Answer 34

<.<

Answer 35

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?

Answer 36

I gave u big O series link...

Answer 37

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

Answer 38

Think ur supposed to use standard format for step representation,

poly, log, quadratic etc...

Answer 39

I don't want to work it all out though...:-)


Wolfram can do it better.

Answer 40

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

Answer 41

Yup!

Answer 42

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

Answer 43

ur welcome.

Answer 44

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)