$ A = \{ 1,2,…,n \} $. Find number of sets $\{B,C\}$ for which $B,C \subseteq A$ , $ |B|=k $ and $ B \cap C=∅ $. Problem: $  A = \{ 1,2,…,n \} $. Find number of sets $\{B,C\}$ for which $B,C \subseteq A$  ,  $ |B|=k $ and $ B \cap C=∅ $.

Answer from solutions (I wrote it as is, there are errors in it though and it's not entirely clear ):
Notice that for every ordered pair $ \langle B,C \rangle $ corresponds the set $ \{ B, C \} $. But there are number of cases for which for two different ordered pairs $ \langle B,C \rangle $  and $ \langle B',C' \rangle $  the set $ \{  B,C \} $ corresponds, and this can occur only if $ \langle B,C \rangle $ and $ \langle B',C' \rangle = \langle C,B \rangle  $ are the different pairs.
These pairs appear only when $ |B|=|C| = k  $ and $ B \neq C $.
Hence, the number of sets $ \langle B ,C \rangle $ of this kind is equal to $ { n \choose k } { n-k \choose k } $.
( $ { n \choose k } $ as the number of options to choose set $ B $ with $ k $ elements out of $ A $  and  $ { n-k \choose k } $ as the number of options to create  set $ C $ with $k $ elements chosen  out of the $ n-k $ elements left from choosing creating $ B $ )
So the number of sets $ \{ B,C  \} $ corresponding the these ordered pairs are $ \frac{1}{2} { n \choose k }   {n-k  \choose  k }   $, In summary the number of sets $ \{ B, C \} $ that satisfy the conditions of the problem is:
$ { n \choose k } \cdot 2^{n-k} - \frac{1}{2} { n \choose k }   {n-k  \choose  k } $

My attempt to solve the problem: We'll find the number of pairs $\langle B,C \rangle $ for which $ B,C \subseteq A$  ,  $~~ |B|=k $ and $ B \cap C=∅ $. The answer is $ { n \choose k } \cdot 2^{n-k} $. To find the number of sets $ \{ B,C \} $ as in the original question, we'll divide the result by 2 since we don't want to count the pairs $ \langle B,C \rangle $ , $ \langle C,B \rangle $ but we only want to count one of them, so the number of  sets $ \{ B,C \} $ as wanted is  $ \frac{1}{2} { n \choose k } \cdot 2^{n-k} $.
Note: The following question may help, I relied on it.
$A= \{1,2,...,n\} $. Find number of pairs $ \langle B,C \rangle $ for which $ B,C \subseteq A ~$ , $~ |B| = k $ and $ B \cap C = \emptyset $
Question: I'm not sure if I'm right but I don't understand the solution either ( it's confusing, and maybe the proposed solution is wrong? ), can you please help? how would you approach and solve the problem?
Thanks in advance for any help!
 A: The shepherd's principle:
In a field of sheep, a shepherd can clearly see and count all of the legs of the animals but the heads and bodies of the sheep blur together and can't easily be distinguished or counted (they're too fluffy).  To count how many animals there are, the shepherd may instead count the legs and then divide the total number of legs by four.
In math, we do the same thing.  If we have a scenario we wish to count and we know we overcounted and had inadvertently counted every item exactly $k$ times... we may divide the total count (including the overcount) by $k$ to get a corrected count.
That only works however in the event that every item was counted $k$ times.
Go back to our shepherd and suppose there were some special two-legged sheep as well.  He won't be sure how many animals there are simply by counting the legs and dividing by four... those two-legged sheep will have only contributed half of an animal to his final corrected count.
In the same way, you dividing your count by $\frac{1}{2}$ is helping to correct those outcomes which you inadvertently counted twice in the previous problem... but you have accidentally caused those outcomes which were correctly counted only once in the previous problem which should also be counted once in the current problem in half when they shouldn't have been.
In the posted solution... they discuss which outcomes were counted twice in the previous problem which we wanted to only count once and so need to correct... and how these compare against those outcomes which were correctly only counted once.  Namely, those outcomes which were pairs $(B,C)$ which have both $|B|$ and $|C|$ equal to $k$, versus those pairs $(B,C)$ which have $|C|\neq k$
