<s>[INST] Finding the number of ways of choosing 4 distinct integers from first n natural numbers so that no two are consecutive. 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.
 [/INST] You must have done something wrong, since for $n=7$ your expression is $35-60+20-4=-9.$ 
To use inclusion-exclusion, your events should be something like "includes $i$ and $i+1$". Now the number of choices which include $i$ and $i+1$ is $\binom{n-2}{2}$, and there are $n-1$ possible $i$, so you should start $$\binom{n}{4}-(n-1)\binom{n-2}{2}...$$ 
Then how many ways are there to have $i$ and $i+1$ and $j$ and $j+1$, for $i<j$? If $j=i+1$ there are $n-3$ ways (and there are $n-2$ choices for $i$). If not there is $1$ way, and $\binom{n-1}{2}-(n-2)=\binom{n-1}2$ choices for $i,j$ (the number of pairs minus the number where $i,j$ are consecutive). So it should continue
$$\binom{n}{4}-(n-1)\binom{n-2}{2}+(n-3)(n-2)+\binom{n-2}2...$$
For the final term, the only way three events can happen is if $i+1=j,j+1=k$, for which there are $n-3$ options. So overall you get
$$\binom{n}{4}-(n-1)\binom{n-2}{2}+(n-3)(n-2)+\binom{n-2}2-(n-3).$$
[This is very close to what you had, the only difference is the extra term for having two separate consecutive pairs of $+\binom{n-2}2$.]
To see that the above is the same as $\binom{n-3}{4}$, note that $(n-1)\binom{n-2}2=3\binom{n-1}3$. Also $(n-3)(n-2)=2\binom{n-2}{2}$, so this can be written as
$$\binom{n}4-\binom31\binom{n-1}3+\binom32\binom{n-2}2-\binom33\binom{n-3}1.$$
This is precisely the inclusion-exclusion formula for the number of choices of $4$ numbers from the first $n$ which don't include any of the last $3$, so is $\binom{n-3}4$.
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