Proof about counting sub-sets (combinatorics) 
Let $Ω_{n} = \{1, 2,\dots, n\}$ and for $0 \leq k \leq n$ let $Ω^{k}$
  be the collection of $k$-element subsets of $Ω_{n}$.  

  
*Define the number $S(n,k)$ as cardinality of $Ω^{k}_n$.  
  
*Prove that $S(n,k) = S(n,n−k)$
  
  
  (Hint: Find a bijection from $Ω^{k}_n$ and $Ω^{n−k}_n$.)

I am kind of new in the world of finding bijections between two sets. I would be glad if someone explains me the general idea to find bijections between sets.
Approach  
Proof
Let's consider the example in which $n=5$ and $k=1$  
\begin{align*}
\Omega^1 &\leftrightarrow \Omega^{5-1} \\
\{1\} &\leftrightarrow \{2, 3, 4, 5\} \\
\{2\} &\leftrightarrow \{1, 3, 4, 5\} \\
\{3\} &\leftrightarrow \{1, 2, 4, 5\} \\
\{4\} &\leftrightarrow \{1, 2, 3, 5\} \\
\{5\} &\leftrightarrow \{1, 2, 3, 4\} \\
\end{align*}
It can be seen in this example that we can establish a relation between $\Omega^{k}$ and $\Omega^{n-k}$. An element $X \in \Omega^{k}$ can be matched up with an element $ Y \in \Omega^{n-k}$ that doesn't contain any element of A. For example, we can imagine that we have n objects. We put k of this n objects in a bucket X and the other n-k objects in another bucket Y. From this example, we can obviously conclude that the X bucket is an element of $\Omega^{k}$ and the Y bucket is an element of $\Omega^{n-k}$. Let's denote the previous relation as $f: \Omega^{k} \rightarrow  \Omega^{n-k}$.  
Now we need to show that $f$ is bijective and to show this we will show it's one to one and onto.  
Assume $f$ is not one to one then there exists at least two elements in $\Omega^{k}$  such that they match the same element in  $\Omega^{n-k}$. Let x be the first element in $\Omega^{k}$ and y the second element in $\Omega^{k}$, so there exists an element $z \in \Omega^{n-k}$ such that $(x,z)\in f$ and $(y,z) \in f$. As we know x and y are sets and they have to differ at least by one element, otherwise they are equal. Let call this element p. So if $p \not\in x$ then $p \in z$, but $p\in y$ which contradicts $f$. Therefore, $f$ is one to one.  
Assume $f$ is not onto then there exists an element $b \in \Omega^{n-k}$ that is not mapped by any element in $\Omega^{k}$.  b is supposed to be mapped by some subset of k elements and this would imply that such subset doesn't exist but we know that it does hence we ensured that $\Omega^{k}$ covers every subset with k elements, so contradiction. This implies that $f$ is onto.
 A: Here’s a hint: let $n = 5$. What’s the relation between these pairs of subsets?
\begin{align*}
\Omega^1 &\leftrightarrow \Omega^{5-1} \\
\hline
\{1\} &\leftrightarrow \{2, 3, 4, 5\} \\
\{2\} &\leftrightarrow \{1, 3, 4, 5\} \\
\{3\} &\leftrightarrow \{1, 2, 4, 5\} \\
\{4\} &\leftrightarrow \{1, 2, 3, 5\} \\
\{5\} &\leftrightarrow \{1, 2, 3, 4\} \\
\end{align*}
Can you do the same thing for $k \neq 1$?
A: Let $f$ be the function from $\Omega_n^k$ to $\Omega_n^{n-k}$ defined by $f: A \mapsto A^c$, where $A^c$ is the complement of $A$ in $\Omega_n$. (Recall that the complement of a set $X$ in a universe $U$ is defined to be the set of all elements in $U$ which are not in $X$.) It can be shown that the function $f$ is one-to-one and onto. Hence, the cardinality of the domain and of the codomain are equal, and in fact are equal to the binomial coefficient $n \choose k$. 
The cardinality of the set $\Omega_n^k$ of all $k$-subsets of $\{1,\ldots,n\}$ is equal to the number of ways to choose $k$ elements from $n$ elements, which is ${n \choose k}$.  This number is usually denoted by $C(n,k)$ rather than $S(n,k)$.  Note that $S(n,k)$ usually denotes the Stirling numbers of the second kind, which is the number of ways to partition an $n$ element set into $k$ nonempty sets.
