How many pairs of students (with one working alone) are possible if there are $21$ students in a class? My solution was to just add up binomials where you choose $2$ people from a smaller and smaller pool of people (since with every choice $2$ people become unavailable): $$\sum_{k=1}^{10} {{2k+1}\choose2}$$ was my answer (it evaluates to $825$). 
But according to my textbook the correct answer is $21! / (2^{10} * 10!)$, which is far from $825$... not only do I have no idea why, but I can't see what's wrong with my answer. 
Just for reference, the question in full is: There are $21$ students in a biology class. The students must pair up to work as lab partners, but, of course, one student will be left over to work alone. In how many ways can the students be paired up?"
Any help is appreciated. 
 A: I dislike both answers.  First, I will make the assumption that the specific order in which the groups are chosen does not matter.  Next, I will make the assumption that the order of the people within the groups does not matter.
Now... Let us begin by choosing which of the $21$ students will work alone.  The poor guy... $21$ options.
Next, among those students that remain, there is one who appears first alphabetically in the roster.  Choose who will be his/her partner.  $19$ choices.  Remove both from the pool of available students remaining.
Next, among those students that remain, there is one who appears first alphabetically in the roster.  Choose who will be his/her partner.  $17$ choices.  Remove both from the pool of available students remaining.
Continue in this fashion, pairing off the students.
This leads to an answer of $21!! = 21\cdot 19\cdot 17\cdots 3\cdot 1 = 13749310575$, exactly the same as the textbook's answer, but written in a different way and avoiding the "division by symmetry" style argument which confuses so many students.
As for what you did wrong... for starters, you summed instead of multiplied.  Next, even if you were to change this into multiplication, you will have incorrectly applied some sort of importance to the order in which the groups were selected.  Correcting your approach by "forgetting the order" that you picked them in, you would have an answer of $\frac{1}{10!}\prod\limits_{k=1}^{10}\binom{2k+1}{2}$ which equals the same as found by the other methods.
A: First of all, you should be multiplying, not adding.  If an event can happen in $n$ ways, and second event can happen in $m$ ways, the two events can happen in $nm$ ways, right?  
That alone will not fix the problem, because of double counting.  Picking Alice and Bob as the first pair is the same as picking Alice and Bob as the fourteenth pair.
Here's one way to do it.  Line up the students in row.  There are $21$ factorial ways to do this.  Then pair up the first two students, the next two students, and so on.  The last student in the row works alone.
Now we have to worry about double counting.  We don't care what order the ten pairs come in, just which pairs we have, so we have to divide by $10!$  Also, in each pair, we don't care which student was first in the line, so each pair has been counted twice.  Altogether, we get $${21!\over10!2^{10}}$$
