Intution for when to divide a product of binomial coefficients by some factor to avoid double counting I'm trying to better understand the intuition for when to divide a product of binomial coefficients to avoid double counting.
For instance, here is an example in which it is appropriate to divide by some number to avoid double counting:
How many ways can we form $2$ pairs of students from $4$ students?
My answer: $$\frac{\binom{4}{2}\cdot\binom{2}{2}}{2!}=3.$$
But in this next example we again look at the product of binomial coefficients but do not need to divide by some number to avoid double counting:
How many ways can we form a five card hand in a game of poker such that there are two red cards and three black cards?
My answer: $$\binom{26}{2}\cdot\binom{26}{3}.$$
Now I think it's apparent in this second example that we are first forming a set of two red cards and then appending to it a set of three black cards, but how does this not over count the converse procedure of appending to three black cards two red cards?
 A: When you pick up a "set" of $r$ objects (at once) from a given collection of $n$ objects, you say there are ${n\choose r}$ possible ways to denote the set of these $r$ objects that you can choose. Observe that, since we are talking about sets here, the order in which the elements appear doesn't matter. All we want is $r$ distinct objects.
That is to say, $\{2,1\} = \{1,2\}$
So in the first example, you have 4 students. You have to choose a pair. So total number of ways of selecting a pair of students is ${4 \choose 2} = 6$
Another way of thinking of this problem can be like this: for the first student you have $4$ choices of total students to start with but after having one student chosen you remain with only $3$ different choices of students and hence the number of ways equals $4*3 = 12$ But this time, let's say students were $a,b,c,d$ then observe that you took both $(a,b)$ and $(b,a)$ as different pairs and hence you counted each pair twice and that's why you need to divide by $2$
In the second example which you stated, again it doesn't matters if we first pick up red cards or black cards, all we want is a set of 5 cards in which 2 are red and 3 are black. There are $26$ red and$26$ black cards so total ways of selecting $2$ red and $3$ black cards is ${26 \choose 2}{26 \choose 3}$. As above, a different argument for counting goes as: There are $26$ choices for first red card and $25$ choices for second red card. SImilarly, $26$, $25$, and $24$ choices for 3 black cards respectively. Hence total number of ways is $(26*25)(26*25*24)$. Again, we made the same mistake. We took $(r_1,r_2)$ pair of red card to be different from $(r_2,r_1)$ and similarly for black cards also we took $(b_1,b_2,b_3)$ different from $(b_1,b_3,b_2)$ different from $(b_2,b_1,b_3)$ and so on. So each pair of red card counted twice and each triple of black cards counted $3! = 6$ times. So we divide by a factor of $2*3!$ and we get the answer as $$\frac{26.25}{2}\frac{26.25.24}{3!} = \binom{26}{2}\binom{26}{3}$$
To add, order matters. So let's say, you have $n$ objects and you carry out the following experiment: You pick an object, take a note of it and remove it from the original collection to start with. You repeat this $r$ number of times. Total number of ways selecting $r$ objects remains ${n \choose r}$ but now since the order matters so there are $r!$ ways of rearranging the objects and hence total number of outcomes is $r!{n\choose r} = \frac{n!}{(n-r)!}$. Again, you can count this in a similar other way as we did in examples. This is called sampling without replacement.
