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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.

RK01
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    When you say that the terms you pick are nonconsecutive, do you mean that no two are consecutive or that they're picked in order so that no consecutive terms appear in order? – Michael Burr Jul 02 '15 at 18:14
  • It means that no two consecutive terms every appear. – RK01 Jul 02 '15 at 18:15
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    Write down $n-k$ circles like this $\bigcirc\quad\bigcirc\quad\bigcirc\dots$. where the numbers not chosen will go, in order. These determine $n-k+1$ "gaps" ($n-k-1$ real gaps, and $2$ endgaps) of which we must choose $k$. – André Nicolas Jul 02 '15 at 18:17
  • 'endgaps' being treated the same – miniparser Jul 02 '15 at 21:32

4 Answers4

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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.

user84413
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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

Michael Burr
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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 ${n-k+1}\choose{k}$

miniparser
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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.

Sisi
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