Equality of Two Permutations

Two permutations $f$ and $g$ of degree $n$ are said to be equal if we have $f\left( a \right) = g\left( a \right)$, $\forall a \in S$.

Example:

If $f = \left( {\begin{array}{*{20}{c}}1&2&3&4 \\ 2&3&4&1 \end{array}} \right)$ and $g = \left( {\begin{array}{*{20}{c}}1&4&1&3 \\ 3&1&2&4 \end{array}} \right)$ are two permutations of degree 4, then we have$f = g$. Here we see that both $f$ and $g$ replace 1 by 2, 2 by 3, 3 by 4, and 4 by 1.

If $f = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}& \cdots &{{a_n}} \\ {{b_1}}&{{b_2}}&{{b_3}}& \cdots &{{b_n}} \end{array}} \right)$ is a permutation of degree $n$, we can write it in several ways. The interchange of columns will not change the permutation. Thus, we can write:

If $f = \left( {\begin{array}{*{20}{c}}{{a_2}}&{{a_3}}&{{a_1}}& \cdots &{{a_n}} \\ {{b_2}}&{{b_3}}&{{b_1}}& \cdots &{{b_n}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{a_n}}&{{a_1}}& \cdots &{{a_n} – 1} \\ {{b_n}}&{{b_1}}& \cdots &{{b_n} – 1} \end{array}} \right)$ then

$f = \left( {\begin{array}{*{20}{c}}{{a_n} – 1}&{{a_n}}&{{a_n} – 2}& \cdots &{{a_1}} \\ {{b_n} – 1}&{{b_n}}&{{b_n} – 2}& \cdots &{{b_1}} \end{array}} \right)$

Therefore, if $f$ and $g$ are two permutations of the same elements of degree $n$, then it is always possible to write $g$ in such a way that the first row of $g$ coincides with the second row of $f$.

Total Number of Distinct Permutations of Degree $n$

If $S$ is a finite set having $n$ distinct elements, then we shall have $n{!}$ distinct arrangements of the elements of $S$. Therefore there will be $n{!}$ distinct permutations of degree $n$ if ${P_n}$ is the set consisting of all permutations of degree $n$. If ${P_n}$ is the set containing of all permutations of degree $n$ then the set ${P_n}$ will have $n{!}$ distinct elements.

This set ${P_n}$ is called the symmetric set of permutations of degree $n$. Sometimes it is also denoted by ${S_n}$.

Thus, ${P_n} =$($f:f$ is a permutation of degree $n$).

The set${P_3}$ of all permutation of degree 3 will have 3!, i.e., 6 elements. Obviously:

${P_3} = \left\{ {\left( {\begin{array}{*{20}{c}}1&2&3 \\ 1&2&3 \end{array}} \right), \left( {\begin{array}{*{20}{c}}1&2&3 \\ 3&1&2 \end{array}} \right),\left( {\begin{array}{*{20}{c}}1&2&3 \\ 2&3&1 \end{array}} \right),\left( {\begin{array}{*{20}{c}}1&2&3 \\ 3&2&1 \end{array}} \right),\left( {\begin{array}{*{20}{c}}1&2&3 \\ 1&3&2 \end{array}} \right),\left( {\begin{array}{*{20}{c}}1&2&3 \\ 2&1&3 \end{array}} \right)} \right\}$