# Thread: 10^80!

1. ## 10^80!

I understand that factorials are more of a pre-algebra subject, but with a number such as 10^80, I believe it is a little more "Higher Level" than that.
I also don't feel like taking all eternity to do the math.

Why 10^80?
10^80 is the accepted number of elementary particles in the observable universe. Well, between the acceptable estimated range of 10^78 and 10^82.

Although the method is flawed, because there is empty space where particles could be, it is the closest example of the chance of every particle being where it is that I, a high school non-astronomer, can come up with.

I hope you can help me, this is the last step in me finishing my script.

Also, if you have a better method of figuring this out, I would be overjoyed to hear it.

Sincerely,
Jares

2. ## Re: 10^80!

is there a question in there?

3. ## Re: 10^80!

First of all, there's not really a question here. From what I gather, you are trying to evaluate either \displaystyle \begin{align*} \left( 10^{80} \right) ! \end{align*} or \displaystyle \begin{align*} 10^{80!} \end{align*}. I can't really tell which it is that you want, and in either case, it would be enormously difficult due to the number of digits it would require. I doubt even supercomputers could evaluate either of these...

4. ## Re: 10^80!

Originally Posted by Prove It
First of all, there's not really a question here. From what I gather, you are trying to evaluate either \displaystyle \begin{align*} \left( 10^{80} \right) ! \end{align*} or \displaystyle \begin{align*} 10^{80!} \end{align*}. I can't really tell which it is that you want, and in either case, it would be enormously difficult due to the number of digits it would require. I doubt even supercomputers could evaluate either of these...
$10^{80!}$ is easy enough.

It's a 1 with

71569457046263802294811533723186532165584657342365 75257710944505822703925548014884266894486728081408 0000000000000000000

zeros after it.

This number of zeros does happen to be 38 orders of magnitude larger than the number of atoms in the observable universe so even if you could print a zero on a single atom the universe has way way too few of them to print out this number.

5. ## Re: 10^80!

I think the OP's point is: if there are $\small 10^{80}$ particles in the observable universe, then the probability that every one is where it is would be $\small \frac 1 {(10^{80})!}$ Not sure what his question is, although clearly if there are only $\small 10^{80}$ particles in the universe it's impossible to actually write out that large a number.

To the OP - be careful with this, because given X particles and X! possible ways to arrange those particles it does not necessarily mean that the one arrangement that actually exists is a long shot. Think about this - if you thoroughly shuffle a deck of cards there are 52! arrangements that the deck can end up in. That's a huge number - but that does not mean that the arrangement you actually get when you shuffle a deck is somehow "special." Only when you pre-define a particular arrangement of cards, then shuffle the deck and get that one arrangement is it remarkable.

6. ## Re: 10^80!

You could try using Stirling's Approximation
Stirling's approximation - Wikipedia, the free encyclopedia

$n! \approx \sqrt{2 \pi n} \left(\frac {n}{e}\right\)^n$

As n tends towrds infinity (and infinite seems like a reasonable estimate for 10^80) this equation becomes

$n! = \sqrt{2 \pi n} \left(\frac {n}{e}\right\)^n$

so

$(10^{80})! = \sqrt{2 \pi 10^{80}} \left(\frac {10^{80}}{e}\right\)^{10^{80}}$
Or
$(10^{80})! = 10^{40}\sqrt{2 \pi } \left(\frac {10^{80}}{e}\right\)^{10^{80}}$
Or
$(10^{80})! = 10^{40}\sqrt{2 \pi } \left(\frac{10}{e}10^{79}\right\)^{10^{80}}$
Or
$(10^{80})! = 10^{40}\sqrt{2 \pi } *3.678^{10^{80}}\left(10^{79}\right\)^{10^{80}}$
Or
$(10^{80})! = 10^\left({{79^{10}}^{80}+40}\right)}\sqrt{2 \pi } *3.678^{10^{80}$

Quite big.

7. ## Re: 10^80!

Originally Posted by ebaines
I think the OP's point is: if there are $\small 10^{80}$ particles in the observable universe, then the probability that every one is where it is would be $\small \frac 1 {(10^{80})!}$ Not sure what his question is, although clearly if there are only $\small 10^{80}$ particles in the universe it's impossible to actually write out that large a number.
The probability that every particle is where it is is 1 because it is there!

To the OP - be careful with this, because given X particles and X! possible ways to arrange those particles it does not necessarily mean that the one arrangement that actually exists is a long shot. Think about this - if you thoroughly shuffle a deck of cards there are 52! arrangements that the deck can end up in. That's a huge number - but that does not mean that the arrangement you actually get when you shuffle a deck is somehow "special." Only when you pre-define a particular arrangement of cards, then shuffle the deck and get that one arrangement is it remarkable.