Combinatorial argument for why $C(n, k) = C(n, n - k)$? 
Give a combinatorial argument to show that 
  $$C(n, k) = C(n, n - k).$$

Any help please?

How many ways are there to select $12$ people from a class of $35$ to give a prize to? We choose $12$ people, and tell them, you are winners! There are $\binom{35}{12}$ ways to do this.
But alternately, we could choose $23$ people from the $35$ and tell them, you are the people chosen not to get a prize. There are $\binom{35}{23}$ ways to do this.  It is clear that there are exactly as many ways to pick the $12$ winners as there are to pick the $23$ losers. 
