Proof for number of ways to select k non-consecutive elements from n consecutive terms. While studying, I found a formula that found the number of ways to select k non-consecutive elements from n consecutive terms, not necessarily the first n consecutive terms, but any n consecutive terms. 
The formula was:   Number of Ways = (n-k+1) C (k)
The C in the formula stood for combinations.
The book gave this formula but no proof for it. I tried to search it up but there were no detailed explanations of the proof.
I'd be glad of assistance if anyone could show me how to prove this formula.
 A: I like the idea presented by Andre Nicolas, but another method would be to line up k dividers (representing the k integers chosen) and let $x_1,\cdots, x_{k+1}$ represent the number of integers in the gaps between the dividers, where $x_1+\cdots+x_{k+1}=n-k\;\;$ and $x_i\ge1$ for $2\le i\le k$ and $x_1\ge0,x_{k+1}\ge0$.
If we let $z_1=x_1+1$, $z_{k+1}=x_{k+1}+1$, and $z_i=x_i$ for $2\le i\le k$, then we have
$z_1+\cdots+z_{k+1}=n-k+2$ where $z_i\ge1$ for each $i$, and this equation has $\dbinom{n-k+1}{k}$ solutions. 
A: You can look at this as the set of integers from $1$ to $n$.  Either you take the first integer or you don't.  
If you take the first integer, then you can't take the second integer.  This means that you have to choose $k-1$ objects from a list of length $n-2$.  
If you don't take the first integer, then you have to choose $k$ objects from $n-1$ consecutive integers.
Then, you can use induction to prove this result.
This is equivalent to the kings problem in one dimension
A: IMO this is the clearest way to solve this.  this is a problem wherein for any given combination you need to $1)$ first 'create' the combination, and $2)$ then label/number the elements.
e.g., you have $5$ consecutive numbers, choosing $2$ non-consecutive.  this means you will have $3$ numbers not chosen, creating four 'slots' where the $2$ non-consecutive can by 'placed.'  however, you need to hold off on numbering them until after they are arranged.  otherwise it doesn't work/make sense.
so for this case have:
$\space\space\space*\space\space\space|\space\space\space*\space\space\space|\space\space\space*\space\space\space|\space\space\space*$
where $|'s$ represent numbers that are not chosen, which are not determined until after the $2$ non-consecutive numbers are placed.  so if choose $1$ and $3$ would be something like:
$\space\space\space\bigcirc\space\space\space|\space\space\space\bigcirc\space\space\space|\space\space\space*\space\space\space|\space\space\space*$
and then afterwards number them.
$\space\space\space\bigcirc_1\space\space\space|_2\space\space\space\bigcirc_3\space\space\space|_4\space\space\space*\space\space\space|_5\space\space\space*$
from here it is clear that in this case solution is ${5-2+1}\choose{2}$ and in general solution is ${n-k+1}\choose{k}$
A: It is equivalent to select k elements from (n-k+1) terms. Then add a term after each of the (k-1) first elements, in order to make the chosen elements non-consecutive. 
