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?
 A: Choose $k$ objects from a set of $n$ and remove them from the set; you can just as well view this as having chosen the $n-k$ objects that you left behind.
If you want to be more technical about it, let $\mathscr{K}$ be the set of all $k$-element subsets of $\{1,\dots,n\}$, and let $\mathscr{C}$ be the set of all $(n-k)$-element subsets of $\{1,\dots,n\}$. The map
$$\varphi:\mathscr{K}\to\mathscr{C}:K\mapsto\{1,\dots,n\}\setminus K$$
is a bijection.
Added: Suppose that $n=5$ and $k=2$. The table below shows the natural pairing of each $2$-element subset of $\{1,2,3,4,5\}$ with its $3$-element complement:
$$\begin{array}{ccc}
\{1,2\}&\longleftrightarrow&\{3,4,5\}\\
\{1,3\}&\longleftrightarrow&\{2,4,5\}\\
\{1,4\}&\longleftrightarrow&\{2,3,5\}\\
\{1,5\}&\longleftrightarrow&\{2,3,4\}\\
\{2,3\}&\longleftrightarrow&\{1,4,5\}\\
\{2,4\}&\longleftrightarrow&\{1,3,5\}\\
\{2,5\}&\longleftrightarrow&\{1,3,4\}\\
\{3,4\}&\longleftrightarrow&\{1,2,5\}\\
\{3,5\}&\longleftrightarrow&\{1,2,4\}\\
\{4,5\}&\longleftrightarrow&\{1,2,3\}
\end{array}$$
When you choose any of the sets in the lefthand column, you’re implicitly also choosing the complementary set in the righthand column: you’re choosing to leave it behind.
A: Here's an easy one. ${n\choose k}$ is the number of subsets of size $k$ from a set of size $n$.  Now look at the unchosen $n-k$ elements.  Each subset of size $k$ generates one subset of size $n-k$. You are done.
A: An idea: each and every set with $\,k\,$ elements ouf of a set with $\,n\ge k\,$ elements determines uniquely a set with $\,n-k\,$ elements, namely: its complement.
A: Both are equal to $\frac{n!}{k!(n-k)!}$ by definition (and the commutativity of multiplication.)
A: 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. 
A: You can visually see it in Pascal triangle which are the values of $n C_k$ where $k=1 \cdots n$
