Probabilities with/without ordering are equal During my math classes, the teacher always highlighted the importance of distinguishing between combinations and permutations. However, I recently noticed that when calculating probabilities with a sample space consisting of equally likely outcomes these two approaches yield the same result.
For instance, let the sample space $U$ consist of $N$ outcomes and consider some subset $A$ of $U$ consisting of $n$ outcomes. We draw $r$ outcomes without replacement from $U$.
Without ordering the probability of all $r$ elements being in $A$ is given by,
$$P(A_\text{no ordering}) = \frac{{n\choose r}}{{N \choose r}} = \frac{n!}{r!(n-r)!}\cdot \frac{r!(N-r)!}{N!} = \frac{n!}{(n-r)!}\cdot \frac{(N-r)!}{N!}$$
With ordering the probability of all $r$ elements being in $A$ is given by,
$$P(A_\text{ordering}) =\frac{n!}{(n-r)!}\cdot \frac{(N-r)!}{N!}$$
Thus the two probabilities are equal.
I am surprised that my teachers haven't pointed this out when studying counting principles and probability. Therefore, I am concerned whether I have misunderstood something or simply made a mistake. I am, of course, aware of the conceptional difference between number of combinations/permutations and that they are related to distinct questions.
I hope someone would take the time to give me their opinion on the matter.
Best regards,
Michael
 A: Yes, this is true. If you're taking $r$ samples from $U$ without replacement, and the event you want to know about doesn't care about their order, then you can use either ordered or unordered sampling.
In general, one or the other might be more straightforward. For example:

*

*Say you are drawing two cards from a deck and you want to know the probability that the first is a spade, and the second is a heart. This talks about a "first card" or "second card", so we have to use ordered sampling: the probability is $\frac{13 \cdot 13}{P_2^{52}}$, where $P^n_k = \frac{n!}{(n-k)!}$.

*Say you are drawing five cards from a deck and you want exactly two out of five to be hearts. You don't care which two are hearts, and don't want to care, so it's easier to use unordered sampling: the probability is $\frac{\binom{13}{2} \binom{39}{3}}{\binom{52}{5}}$.

You can always make a problem into an ordered sampling problem if it isn't already, but you might make the problem more complicated that way. In the second example I gave, we'd have to write $\frac{\binom 52 P^{13}_2 P^{39}_3}{P^{52}_5}$: the extra factor of $\binom 52$ counts the number of ways to pick which two cards are hearts.
But by far the most common mistake students make is to be inconsistent and use ordered sampling for part of the calculation and unordered sampling for the rest. This is always wrong.
