Question

) there are exactly 20 students currently enrolled in a class. how many different ways are there to pair up the 20 students, so that each student is paired with one other student

Answer 1

Assuming that "pair up students" means "divide up all 20 of the students into groups of two," and we regard two pairings as the same if and only if, in each pairing, each student has the same buddy, then I believe that your answer of 20! / [(2!)^10 * (10!)] is correct. (And I also believe that this is the best interpretation of the problem as you've stated it.) 

There are at least two ways to see this (possibly more). 


One way is to note that, first, we have to select 2 students for the first pair; that's C(20, 2) (where by C(20, 2) I mean "20 choose 2"; that is, 20! / (18! * 2!). ) 

Then, for each way of selecting 2 students for the first pair, I have to select 2 of the remaining 18 students for the second pair, so I multiply by C(18, 2). 

Continuing in this manner, I get C(20, 2) * C(18, 2) * ... * C(2, 2). 

But it doesn't matter in this situation the order in which I pick the pairs of students. Since there are 10! different orders in which I could pick the individual pairs, then I want to divide the above by 10!, giving me the answer 

[C(20, 2) * C(18, 2) * ... * C(2, 2)] / 10!. 

This is the same as your answer, because C(n , 2) = n(n - 1) / 2, so we can simplify the above as 

[(20 * 19) / 2 * (18 * 17) / 2 * ... * (2 * 1) / 2] / 10! 
= 20! / [2^10 * 10!] 
= 20! / [(2!)^10 * (10!)]. 



Another way is to reason as follows: 

1. First, arrange the 20 students in a line; there are 20! ways to do this 
2. We can get a pairing by pairing the 1st and 2nd students in line together, the 3rd and 4th students together, etc. 
3. But if I switch the order of the 1st and 2nd student, then this doesn't give a different pairing. I don't want to count both orderings separately, so I divide by 2! 
4. The same argument from step 3 holds for the 3rd and 4th student, the 5th and 6th student, etc., so I need to divide by 2! nine more times 
5. Finally, the particular order in which I selected the ten pairings are unimportant--for example, the following orderings don't produce different pairings: 

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 
3, 4, 1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 

So I need to further divide by the number of ways I can arrange the ten pairs, which is 10!.