Counting number of combinatorial sequences. Let $n$ be a positive natural number. A sequence of $n$ positive positive integers (not necessarily distinct) is called a ”four-group” sequence if it satisfies
the following requirements: for any natural $k \geq 2$, if $k$ occurs in this sequence, then $k − 1$ occurs, and the first position at which the number $k − 1$ occurs,
is earlier than the last position where the number $k$ occurs. Find the number of ”four-group” sequences of length $n$.
 A: 
There are $n!$ four-group sequences of length $n$. 

Here is a bijection between FGS's and permutations. Given a permutation of length $n$, partition it into maximal decreasing consecutive subsequences. For example, with $\pi=[\pi_1,\pi_2,\dots,\pi_{n}]$ in one-line notation, and $n=10$,
$$
[7,5,2,1, 3,10,6,9,8,4]\implies 7,5,2,1\def\div{\;\Big|\;}\div3\div10,6\div9,8,4 
$$
To make the FGS corresponding to this permutation, we place a $1$ at each index in the first subsequence (at spots $7,5,2$ and $1$), we place a $2$ at each index in the second decreasing subsequence (at spot $3$), and in general place the number $k$ at each position appearing in the $k^{th}$ block. The result in this example is
$$
1, 1, 2, 4, 1, 3, 1, 4, 4, 3
$$
Note that the set of numbers appearing will always be a consecutive interval of integers containing $1$. Furthermore, the maximality of the decreasing subsequences implies that the first instance of $k-1$ is before the last instance of $k$, for each $k$. 
The inverse map is as follows. Given an FGS, arrange the indices of the appearances of $1$ in decreasing order, then append the indices of the appearances of $2$ in decreasing order, and so on. I leave it to you to verify these maps are inverse to each other.
