When are permutations and combinations interchangable? I was recently reading through a probability textbook and came across an example in the conditional probability section that went like this, "An urn contains $10$ marbles; $8$ are red and $2$ are white. You choose two marbles at random. Let $\mathcal{R}_1$ be the event that the first marble is red, and let $\mathcal{R}_2$ be the event that the second marble is red. Suppose the first is red, i.e. that $\mathcal{R}_1$ is true. What is the probability that the second is red given this information? i.e what is $P(\mathcal{R}_2|\mathcal{R}_1)$?
And they give the answer as $\frac{P(8,2)/P(10,2)}{8/10}$ and this bothered me because I could not for the life of me convince myself that the order of the chosen, supposedly identical, red marbles mattered, therreby warranting the use of permutations. So I tried calculating it with combinations, i.e $\frac{C(8,2)/C(10,2)}{8/10}$... and got the exact same result!
Now I'm curious. Is there any literature on when one can use either $P$ or $C$ to count the same values? Does this only happen because I am using ratios of $P$'s and/or $C$'s?
If I'm blowing this all out of proportion. I would very much welcome an explanation as to why order and thus permutations are required. Thanks!
 A: The question explicitly states that the FIRST is red, then later that the SECOND is red,  so order is present in the question,  thus using P(n,k) makes sense.
A: Observe that in the compound faction in the fraction in the numerator, numerator and denominator are either both permutations or both combinations.  What matters in computing probabilities in finite sample spaces using the 'basic probility rule' is counting 'simple outcomes' consistently in both the event of interest and the sample space.
In general counting problems (as distinct from counting as a part of elementary probability), use permutations to count things involving a sequence of selections when order of selection matters, combinations when order is unimportant.
Here's an example.  The math club has eight members.  There are
$$C(8,3)=\dfrac {8!}{5!3!}$$
ways to choose a committee of three but there are
$$P(8,3)=\dfrac{8!}{5!}$$
of choosing three members, the first to be president, the second vice president, and the third to be secretary.
A: In this case it doesn't matter which method you use, because the event you are calculating the probability of, "both are red", does not depend on the order. Choosing a first marble, then a second, is the same as first choosing two marbles, then picking which goes first. In this case you know whether both are red after the first step, so the second step is not necessary and you can work with combinations. It is crucial that you are choosing items without replacement for the two methods to be equivalent.
However, for many questions it will make a difference and the answer which takes account of order is the correct one.
