Out of Pod Experience

It is also winter time and some matches may get postponed.

And that's why we have two possible answers, depending on what the OP meant, the context of his question.

If he meant "overall win/loss/draw statistics of a single team playing 10 consecutive matches with doesn't matter who", it's 66 possibilities.
Same answer if the context was "overall number of wins/losses/draws in a round of 10 matches between whatever teams". Still 66.

If he meant "number of possible ways to fill in a sports-betting sheet with 10 specific matches on it", it's 59049 versions.

Those answers, either 66 or 59049 are the only possible answers, depending on context. There are no other possible interpretations of the question the OP put in which the answer would be different from one of those two.

This is turning into another 4/2+2 thread.

I believe the permutations is the multiplication of all the factorals. 3^10 is wrong. My wife explained it to me, I sorta scratched my head, but the gist of it I got was just multiply the factorals of the stuff then divide by something or another.

Edit as a side note doing 3^10 doesn't get you anything meaningful for the most part, it's not the combinations or permutations.

Mnemonics:
Permutations and Powers start with the letter "P", i.e. 3^10
Permutations for when order matters. The letter "p" comes after "o", i.e. ordered.

Combinations use factorials and factorials get big Crazy fast. Crazy and Combinations start with "C", factorials contains the letter "C". (The factorial of 70 is a 101 digit number, whereas 3^70 is just a 34 digit number...)
Combinations for when order doesn't matter. (Go "C"razy with the ordering.)


You can also google "permutations" and "combinations" and find a dozen math tutorials, online examples, and how-to's on the subject.

In fact I know a different Definition.
edit: The Wikipedia article of Permutation uses the same definition as I do, but with a hint on your Definition.

From what I learned a Permutation is the recombination
of the entries a given vector of length n.
Example:
(2,1,3) is a Permutation of (1,2,3)
(3,2,1) is a permutation of (1,2,3)
(1,2,3) is a permutation of (1,2,3)


A combination is a recombination of a subset of k entries from a vector of length n.
Example:
(1,2) is a combination of length 2 from (1,2,3)
(1,1) is a combination of length 2 from (1,2,3)

For combinations there are 4 different possibilities to draw:
With replacement, with consideration of order
Without replacement, with consideration of order
With replacement, without consideration of order
Without replacement, without consideration of order

With replacement means you can draw a number more than once, i.e.
if you first draw the 1 out of (1,2,3) you again draw from (1,2,3) in the second draw

Without replacement means every number can only be drawn once, i.e.
if you first draw the 1 out of (1,2,3) you then draw from (2,3) in the second draw

With and without consideration of order has been explained above.
Without consideration of order means that drawing (2,3) from (1,2,3) is seen as
the same result as drawing (3,2).


Another definition says that a combination is what I explained above, but only
without consideration of order. If you are considering the order it is then called variation.


Regardless of how you call it, the formulas for the number of recombinations for the permutation and 4 combinations are:
(with choose(n,k) = n! / (k! * (n-k)!) and n! = 1 * 2 * ... * n)


Permutation:
n!

Combination with replacement, with consideration of order:
n^k

Combination with replacement, without consideration of order:
choose(n+k-1, k)

Combination without replacement, with consideration of order:
n! / ((n-k)!)

Combination without replacement, without consideration of order:
choose(n,k)



Regardless of how you name it, permutations, combination, variations,... whatever. Basically these are the five formulae you need to do combinatorics (In fact you only need four as a permutation is a combination without replacement, with consideration of order with n=k).



edit2: So what I am basically saying is: Akita is right. Listen to her!
You want to draw k=10 times from n=3 outcomes. You cannot do that without replacement.
Taking order into account you get n^k = 3^10
Not taking order into account gets you choose(n+k-1,k) = choose(12,10) = 12! / (10! * 2!) = 11*12 / 2 = 66

\o/

I think we're both misreading eachother's posts...my bad...

My name is Ralph.