List all the permutations of {1,2,3,4}. Which are even, and which are odd? The answer is:
There are 24 permutations. 
The 12 even permutations are: id , (1 2 3 4) , (1 3 2 4) , (1 4 2 3) , (1 2 3) , (1 2 4) , (1 3 2) , (1 3 4) , (1 4 2) , (1 4 3) , (2 3 4) , (2 4 3).
The 12 odd ones: (1 2) , (1 3) , (1 4) , (2 3) , (2 4) , (3 4) , (1 2 3 4) , (1 2 4 3) , (1 3 2 4) , (1 3 4 2) , (1 4 2 3) , (1 4 3 2).
The question is, why we simply do not write out all the combinations as follows:
(1 2 3 4) , (1 2 4 3) , (1 4 2 3) , ect.?
Also how do we get combinations such as (1 2) , (1 3) , (1 4 3) and so on?
Thanks
 A: It is a matter of cycle notation.
$\{1, 2, 3, 4\}$ is the set.  $(1\; 2\; 3\; 4)$ is a permutation of the set; a sequence of transpositions of that set represented by the rotation of elements.  (A transposition is a permutation resulting from an exchange of just two elements.)
This permutation can also be expressed as: $\begin{pmatrix}1 & 2& 3 &4\\ 2 & 3 & 4 & 1\end{pmatrix}$ (Cauchy notation; initial state above, end state below).
It means that the first element is exchanged with the second, then the third, and finally the fourth.  This is also expressed as a product of the individual transpositions: $(1\; 2\; 3\; 4) = (1\; 2)(1\; 3)(1\; 4)$  That is an odd number of transpositions (three), meaning it has an odd parity.  
$(1 \; 2)$ is a single transposition; so it's an odd parity.  $(1\; 2\; 4)$ is two transpositions; so it's an even parity.
$(1\; 2) \equiv \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4\end{pmatrix}$
$(1\; 2\; 3) = (1\; 2)(1\; 3) \equiv \begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 1 & 4\end{pmatrix}$
