[Combinatorics]
Problem: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.
Compute two other 6-labels with the same 4-signature as 462135.

Solution: $352146,362145,452136,562134$

Final answer: $352146,362145,452136,562134$