$1,2, \ldots, n$ are permuted. None of the numbers $1,2,3$ are adjacent and $n>4$. The numbers $1, 2, \ldots, n$  are permuted. How many different permutations exist such that none of the numbers $1, 2, 3$  are adjacent when $n>4$?
Solution:
$4,5, \ldots, n$ can be shuffled in $(n-3)!$ ways and $3!$ ways to arrange $1,2,3$.  There are $n−2$ slots that are separated by $n−3$ shuffled numbers, and if we insert each of $1,2,3$ into a different slot, they cannot be adjacent. There are $\binom{n - 2}{3}$ ways to do this.
Why is it $\binom{n-2}{3}$ ways? I don't get the explanation.

There is a permutation of $1,2,3$ giving factor $3!$
There is a permutation of $4,5,\dots n$ giving factor $(n-3)!$
The numbers $4,5,\dots,n$ are "separated" (e.g. like $\star7\star5\star4\star6\star$ if $n=7$) and $n-2$ slots/stars exists.
These stars must looked at as candidates for placing there exactly one of the numbers $1,2,3$.
So we must select $3$ of $n-2$ stars giving factor $\binom{n-2}3$.
Then we end up with a total of: $$3!(n-3)!\binom{n-2}3$$possibilities.
