If i have this arrangement $(1,2,..,k,...,n)$. $\sigma$ is a permutation of the $n$ elements and $\rho$ is a permutation of the first $k\leq n$ elements.
How i can show that the following set:
$$\{\sigma \circ \rho \hspace{0.1cm} | \text{$\sigma, \rho$ are permutations}\}$$
has in him $k!$ times each permutation of $n$ elements?.
I tried to confirm that using permutations of $3$ elements $(1,2,3)$ and permutations of $2$ elements $(1,2)$. So:
The permutations of the $3$ elements are $\{(),(23),(123),(12),(132),(13)\}$
The permutations of the $2$ elements are $\{(),(12)\}$
If i do the composite of each one of the permutations of $3$ elements with each one of the permutations of 2-elements i have the set:
$$\{(),(12),(23),(132),(123),(13),(12),(),(132),(23),(13),(123)\}$$
As can be seen, each element of the permutations of $3$ elements are in the set $2!$ times.
There are $k!$ permutations of the first $k$ elements. For each target permutation of the $n$ elements I claim that for each $\sigma$ I can find exactly one $\rho$ that produces the target permutation. If the target permutation is $\kappa$ we want $\kappa = \sigma \circ \rho$, which means $\sigma^{-1} \circ \kappa = \rho$. As there are $k!$ choices for $\sigma$ there will be $k!$ ways of achieving it with $\sigma \circ \rho$