Choosing numbers without consecutive numbers.

Assuming that the order of choice doesn’t matter, imagine marking the positions of the $r$ chosen numbers and leaving blank spaces before, between, and after them for the $n-r$ non-chosen numbers; if $r=3$, for instance, you’d get a skeleton like $_|_|_|_$, where the vertical bars represent the positions in $1,2,\ldots,n$ of the chosen numbers. The remaining $n-r$ numbers must go into the $r+1$ open slots in the diagram, and there must be at least one of them in each of the $r-1$ slots in the middle. After placing one number in each of those slots, we have $n-r-(r-1)=n-2r+1$ numbers left to place arbitrarily in the $r+1$ slots. This is a standard stars-and-bars problem: there are

$$\binom{(n-2r+1)+(r+1)-1}{(r+1)-1}=\binom{n-r+1}r$$

ways to do it. The reasoning behind the formula is reasonably clearly explained at the link.