Question:
Show that the number of ways of choosing 4 distinct integers from the first $n$ numbers so that no two are consecutive is $\binom{n-3}{4}$.
I tried to find it by the principle of inclusion and exclusion. The total number of ways of choosing 4 integers here is $\binom{n}{4}$, number of ways to choose them such that at least 2 are consecutive is $\binom{n-2}{2}×(n-1)$ and at least three are consecutive is $\binom{n-3}{1}×(n-2)$ and all 4 can be consecutive in $n-3$ ways.
Thus total number of ways such that no two are consecutive by principle of inclusion and exclusion should be $$\binom{n}{4} - \binom{n-2}{2}\times(n-1) + \binom{n-3}{1}\times(n-2) - (n-3) $$
How can I show this is $\binom{n-3}{4}$? Or did I do anything wrong? Please help. Thanks in advance.