Math Help

[wolven7]

What is the function/equation for finding how many different permutations of a set are available?

What I mean is, if I have a set of 78 things, how can I easily find how many different ways they can be combined?

Right now all I have is the brute force x=n(o). Meaning that the number of permutations is equal to the number of things in the set times the previous number of permutations. So if there are three things in the set, the number of combinations of those things is equal to three times two. Two being the number of combinations available to a set consisting of two distinct things. And we continue, in that manner. As I said: Brute Force.

So, I'm pretty sure I know there has to be an easier way, but my math, she is not strong.

Help?

Comments

[ladylabyrinth.livejournal.com]

If it's a permutation, and order of items is important, then you'll use n!/(n-r)! where n is your total number of items and r is the number of items per permutation (17 gadgets taken 4 at a time would be 17!/(17-4)!. Oh, and !=factorial, which is nx(n-1)x(n-2)x(n-3)x...x(n-(n-1)). A combination, where order doesn't matter, would be n!/r!x(n-r)!, n and r having the same meanings as in permutations. These are written as ⁿP_r and ⁿC_r, respectively. Also, here, have a link.

[wolven.livejournal.com]

The question is about both. Because, in Tarot, we have a set of 78 cards, taken (usually) 10 at a time.

I also want to know how many different ways I can arrange All 78 of them. For instance, I know that if there were only 12 of them it would be 479,001,600 different arrangements of all 12.

So, if you could give me an example of the function, in action?

[ladylabyrinth.livejournal.com]

does order matter? Is repetition allowed? this effects the total...Here, have a calculator. ^_^

[wolven.livejournal.com]

SWEET! That EXACTLY what i needed! Thank you.

[ladylabyrinth.livejournal.com]

*bows* Glad to help! Now invite me to your next party, or something. I need to GET OUT more.

[wolven.livejournal.com]

You're right; we need to have a party, soon...

[raidingparty.livejournal.com]

Random person you don't know's birthday party on Friday at my place. Possibly followed by "How I Became the Bomb" concert.

[ladylabyrinth.livejournal.com]

If order does matter, and repetition is NOT allowed, then ⁷⁸P₁₀=4.56617696981846e+18, which is a LOT.

If order does NOT matter, and repetition is still not allowed, then it's ⁷⁸C₁₀=1,258,315,963,905

ETA: I'm assuming there's NO repetition, since each card only occurs once, and can only be drawn/placed once, yes?

[wolven.livejournal.com]

Order is important.

So, Precisely.

But I can never remember how to express things multiplied by "e". It always helps me to do these things the long way around.

How do I find out how many zeros that is?

[ladylabyrinth.livejournal.com]

Erm...what I can find on e is that it's irrational and therefore not able to be calculated to any definite number. So, erm...I'm actually not sure! My google-fu has failed me at this point. Ask nice michette person down there for further assistance. ^_^

[davidsfoley.livejournal.com]

Not that I'm an expert by any means, but in I think that the "e" in this case is not the natural logarithmic "e" but instead "E" which is the calculator's way of saying that your answer is in scientific notation form. Thus your first answer is 4.56617696981846 x 10^{^18}. Or 4,566,176,968,184,600,000.

Mind you, this is such a large number that the calculator probably cut off the last five digits; they probably aren't zeroes. My memory is pretty shaky here, but Wikipedia (http://en.wikipedia.org/wiki/Combination) knows how to do it by hand, if you're really interested in finding those extra digits.

[ladylabyrinth.livejournal.com]

I knew there was something wrong with what I was reading about e and the answer we got. Thank you for the clarification!

[wolven.livejournal.com]

That makes a good bit of sense, actually. That's an order of magnitude Higher than I estimated.

[wolven.livejournal.com]

As I say: My math, she is-a no good.

[michette.livejournal.com]

i'm so glad that someone answered this before i read it. thought i was gonna have to pull out the pre calc book, and sincei 'm on antibiotics right now (read: liz is taking stupid pills again) it probably wouldn't have made sense

[wolven.livejournal.com]

Well, I appreciate the thought, anyway :)