Let $n \geq 4$. How many permutations $\pi$ of $[n]$ have the property that $1,\;2,\;3$ appear in the same cycle of $\pi$, while $4$ appears in a different cycle from $1,\;2,\;3$? For instance, when $n = 4$, there are two such permutations: $(123)(4)$ and $(132)(4)$.
When $n = 5$, would the permutations be $(123)(45)$, $(1235)(4)$, $(123)(4)(5)$, etc? Would the formula for $n = 5$ be $\frac{5!}{2!3!}$? What is the general formula for $n$?
Let $1$, $2$, $3$ appear in a cycle of length $k$. Then we can choose the remaining $k-3$ elements of the cycle from $n-4$ elements in $\binom{n-4}{k-3}$ ways and arrange them in a cycle in $(k-1)!$ ways, and the remaining $n-k$ elements can be permuted in any of $(n-k)!$ ways. Thus there are
$$ \binom{n-4}{k-3}(k-1)!(n-k)!=\frac{(n-4)!(k-1)!(n-k)!}{(n-k-1)!(k-3)!}=(n-4)!(n-k)(k-1)(k-2) $$
such permutations. Summing over $k$ yields a total of
$$ (n-4)!\sum_{k=3}^{n-1}(n-k)(k-1)(k-2)=\frac{n!}{12} $$
permutations. The simple result suggests that there should be a more straightforward way to obtain it.