There are $10$ students in the class. How many ways can we split them into groups of size $2$? 
There are $10$ students in the class. How many ways can we split them into groups of size $2$?

My answer is
$$\dbinom{10}2\dbinom82\dbinom62\dbinom42\dbinom22$$
Obviously, there are repetitions between the combinations as the order of the groups doesn't matter. I am trying to figure out a good explanation for the correction. What is a general solution for a group of $K$ from $N$?
 A: You should provide the logic behind your answer.  It appears you are choosing $2$ for the first group, then $2$ of the rest for the second group, and so on.  The error is that you can reorder the groups any way you want.  Given one partition into pairs, you could select the third pair first, the fourth pair second, and so on and get the same partition.  There are $5!=120$ ways to order the pairs, so you need to divide your answer by $5!$ because you have counted each partition that many ways.
A: When you count something, one general technique is to over-count and then divide by the number of times each unique solution appears. So you should form a fraction $A/B$ where you count a set of $A$ things and each solution appears $B$ times.
So look at the numerator $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$. Can you explain what this counts? Can you figure out how many times each set of $5$ pairs is included in this number? The number $5!$ has been thrown around. Try to figure out what $5!$ counts in relation to the problem.
The general solution is given by a double factorial. Namely, it is
$$ 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = (10 - 1)!! = \frac{10!}{5!2^5}. $$
If there are $2n$ people forming $n$ pairs, then the answer is
$$ (2n - 1)!! = \frac{(2n)!}{n!2^n}. $$
Try to figure out how the first formula simplifies to this. I.e. show that
$$ \frac{1}{n!}\binom{2n}{2}\binom{2n-2}{2}\cdots \binom{4}{2}\binom{2}{2} = \frac{(2n)!}{n!2^n}. $$
Also try to explain combinatorially how you get $(2n)!/n!2^n$ as the solution. (Hint: think about what $(2n)!, n!$ and $2^n$ could mean in terms of the $2n$ people and the $n$ pairs.)
Then as a bonus, explain combinatorially why $$(2n - 1)!! = (2n - 1)(2n - 3)(2n-5) \cdots 5 \cdot 3 \cdot 1$$ is also the solution.
Then once you've done all that, you should be able to answer your general question: how many ways are there to partition a set of size $kn$ into sets of size $k$.
A: A possible way is to consider your question as follows where $K|N \Leftrightarrow N = K\cdot r$


*

*all possible arrangements of $N = 10$ items: $\color{blue}{N!}$
Now, divide the places for  the arrangements in $r=5$ blocks of $K=2$ consecutive items:


*

*$\underbrace{(1 \ldots K)(K+1 \; \ldots 2K) \ldots (N-K+ 1 \; \ldots N)}_{r \: blocks = r\: groups}$
So, you can see that a given grouping into groups of size $K$ is achieved by several arrangements as follows:


*

*Rearranging the $r$ groups among each other and rearranging the items within the groups of $K$ items: $\color{blue}{r! \cdot (K!)^r}$
So, you get as a general formula
$$\frac{\color{blue}{N!}}{\color{blue}{r! \cdot (K!)^r}}$$
In your special case you have $N = 10, K = 2, r = 5$:
$$\frac{10!}{5! \cdot (2!)^5} = 945$$
