Equality of Two Permutations

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



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:

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