$1,2, \ldots, n$ 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.

We can count the cases


*

*the permutations are n!


from which we need to eliminate the ways


*

*we can arrange $1,2,3$ adjacent that is: $3!(n-2)(n-3)!$ 

*we can arrange exactly a pair adjacent that is: $3!(n-3)![(n-3)(n-4)+2(n-3)]=3!(n-3)!(n-3)(n-2)$
