Simplifying factorials within combinatorics

Question

Find how many arrangements of $n$ people around a circle are possible if two particular sit apart.

I prefer doing this the complementary way so how many arrangements there are without any restriction $-$ the number of ways of two particular people sitting together

This equates to:

$(n-1)! - (2\times(n-2)!)$

This is right (hopefully) but how do I simplify this to the answer : $(n-3) \times (n-2)!$

Answer 1

$(n-1)! - (2\times(n-2)!)$

= $(n-1)(n-2)! - (2\times(n-2)!)$

Taking (n-2)! common,

= $(n-2)![ n-1-2]$

= $(n-3)(n-2)!$

Answer 2

Expand the $(n-1)!$ by $$\begin {align}(n-1)! - (2(n-2)!)&=(n-1)(n-2)!-2(n-2)!\\ &=(n-3)(n-2)!\end {align}$$

Answer 3

Your answer is correct :

$(n-1)!-2\cdot(n-2)!=(n-2)!\cdot(n-1-2)=(n-2)!\cdot(n-3)$.