1. ## combinatorics ...

Hey guys here's my confusion, any help clearing it up would be wonderful ...

Let's say you've got 2 * 6-sided dice

The sample space is 6^2 = 36 potential outcomes

Q/ Is the sample space equivalent to the number of "permutations"?

I ask because if you labelled the dice A and B, and cared about the throwing order, you'd actually have 72 potential outcomes in total....

Or let's say you've got 2 fair coins:

Sample space = 2^2 = 4

And for combinations HT = TH
For permutations HT distinct from TH

But again, if you labelled the coins A and B, you'd have 2(2^2) total possible outcomes

So is the definition of permutation at the objection selection level AB vs BA? OR is it that once selection's made, do each of HH HT TH TT represent a distinct permutation out of the 4 possible.

Thanks in advance for clearing this up.

2. ## Re: combinatorics ...

The sample space is by definition the set of all possible outcomes of your experiment. In the case of an experiment with two coins, you have four possible outcomes {TT, TH, HT, HH}. When you talk about permutations, you need to be clear about what it is that is being combined. For n possible outcomes, there are n! possible permutations by our counting rules. I'll assume your text adequately explains why that is.

So what are we talking about when we are talking about the coins? The only meaningful thing I can think of is to speak of the faces of the coins. How many faces do we have? There are 4 faces, 2 on each coin. Thus, we have 4! = 24 possible permutations (e.g., there are two pairs HH, when it one coin is in the first position and then when it is in the second). This is clearly not the sample space of interest. This is most obvious by the fact that we don't care which coin it comes from (e.g., HH can only occur once in the sample space, regardless of which coin produces the H; it's enough that both of them did). If this is not correct, then as I said, you need to be clear about what the permutations are of. What is the object being ordered here?

3. ## The Following User Says Thank You to bryangoodrich For This Useful Post:

SBruce (07-22-2012)

4. ## Re: combinatorics ...

Hey Bryan, thanks for taking the time to explain. I think it's mostly cleared it up, but there's still one part I'm not clear on so I'll write what I understand based on what you wrote:

"Sample space" = all possible outcomes of a specific experiment (ordering of trial spaces or run of trials)

whereas "permutations" are the n! ways that the experiment can be run (object ordering)

So the total potential outcomes of all experimental orderings for a bunch of objects = sample space for any given experiment * n! ways experiment can be run.

The above definition seems to make sense, but wouldn't fit what you wrote about 2 coins A,B having 4 faces in total giving 4! permutations... I think?

For instance, imagine four * 6-sided dice, one w/ numbers, other letters, other symbols, other colours:

In one experiment, they're rolled in a particular order NLSC so the sample space for that run is 6^4. In another experiment they're rolled in another order SLNC and the sample space is again 6^4. etc etc. Following this logic, there's 4! ways of ordering the 4 dice into each trial position for any given experiment, and each ordering (experiment) results in 6^4 potential outcomes, so there's 4! * 1296 = 31104 outcomes in total.

If we applied the analogy of the two-coins A,B having 4 combined faces in total and 4! permutations to the above four dice example, hence thought of the four dice as having a combined 6 * 4 = 24 faces this would give 24! potential permutations which is astronomical! nothing like 31103 ??

Apologies if I miscalculated this or misunderstood, I do appreciate the help to sort out whatever's got my thinking stuck.

Actually mate I've actually figured out what was confusing the issue for me...
It was a confusion between object selection order (mutually exclusivity n!/(n-k)!) and the number of values each object can take on once selected (#values^n = sample space).

So Jack, Jill and Jane are in a race ... there's 3! potential 1st, 2nd and 3rd place results (=permutations), or just 1 result if place didn't matter and all completed the run (=combination).

And if just to mess with them, there were 3 distinct "prizes" of \$10, \$20 and \$30, which were randomly but mutually exclusively allocated to each participant independent of the race result, so 3! potential reward scenarios (=permutations) or 1 scenario if they didn't care about the value of the money (=combination)...

So if we did care about the combined effect of race result and reward scenario, there'd be 3! * 3! = 36 potential outcomes of race result * reward result in total yeah?

If so, "permutations" would just describe the potential orderings of the mutually exclusive results for each experimental condition separately (race, or reward), the combined total potential results given by 3(P)3 * 3(P)3 = 36, not (3*3)! = 9! = 362,880 .. ?

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