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How many ways are there to select $k$ elements from the set $[n]$ such that the numbers selected differ by at least three?

I thought of considering two cases: $n-k$ is even or $n-k$ is odd. If $n-k$ is even, then I arrived at ${\frac{n-k}{2}+1}\choose{k}$ ways and if $n-k$ is odd, then there are ${\left\lfloor \frac{n-k}{2} \right\rfloor +1} \choose {k}$ possibilities. To answer this question, I was analysing the ways of choosing k elements from the empty spaces that exist when considering an array on $n-k$ $\textbf{0's}$. Any suggestions for a better argument and what's the answer for a more general case: the k selected numbers should differ by at least $d$ ?

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    $\begingroup$ I'd work recursively. Either the first element isn't in the selection, in which case you are down to selecting $k$ elements from $n-1$ total, or it is in the section, in which case you are selecting $k-1$ from $n-4$. $\endgroup$
    – lulu
    Oct 21, 2023 at 17:07
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    $\begingroup$ When you say differ by at least three, I am taking it that $(1,3)$, say, is unsuitable, but $(1,4)$ is ? $\endgroup$ Oct 21, 2023 at 20:46

2 Answers 2

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I had mistakenly the difference as $2$ instead of $3$, amended

  • For a difference of $3$, after every chosen number (bullet) except the last , there must be at least two spacers (circles), $\boxed{\bullet\circ\circ}\circ\circ\circ\boxed{\bullet\circ\circ}\bullet$

  • From $n$ unnumbered tokens, take out $2(k-1),\;\; (n-2k+2)$ remain

  • Choose any $k$ from the above in $\binom{n-2k+2}{k}$ ways, mark them, but don't number them yet.

  • Insert from the taken out tokens, $2$ immediately after each of the chosen tokens (except the last), and now allot serial numbers

  • This generalises to $\dbinom{n-(d-1)(k-1)}{k}$


First Approach added back (simpler?)

  • Add $2$ elements [total now $(n+2)$] and form $k$ blocks of one chosen and two unchosen $\boxed{\bullet\circ\circ}$

  • There are now $(n+2-2k)$ entities $(k$ blocks plus $(n+2-3k)$ "singles")

  • Place the blocks in $\binom{n+2-2k}{k}$ ways and discard the last $2$ elements

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  • $\begingroup$ I like your argument, but I think you've under-counted! These types of blocks are allowed to be next to each other, which you aren't accounting for. You can fix it by using blocks of width $2$ instead. You should then only add one number at the end, but there are $(n + 1 - 2k) + 1$ spaces in between numbers to place a block, so our answers agree. (Sanity check: $k = 1$ should give $n$ sets) $\endgroup$ Oct 21, 2023 at 17:40
  • $\begingroup$ @IzaakvanDongen: I had somehow misread the minimum differemce as $2$ instead of $3$, have amended, thanks. $\endgroup$ Oct 21, 2023 at 18:38
  • $\begingroup$ To me "difference of at least $3$" means eg $\{1, 4\}$ is allowed, but $\{1, 3\}$ isn't. I actually thought your previous answer was closer! Am I right that your result is there's $\binom{n+3-4k}k$ such sets? I think in the extreme case $k = 1$, the answer should be $n$, and when $n=3k-2$, the answer should be $1$. This seems incompatible with your result! As I understand it, when you "place $k$ blocks in $\binom{n+3-4k}k$ ways", you are choosing $k$ gaps in between the remaining numbers. I thought the problem was that this didn't allow $\boxed{\bullet\circ\circ}\boxed{\bullet\circ\circ}$. $\endgroup$ Oct 21, 2023 at 18:45
  • $\begingroup$ @IzaakvanDongen: Thanks, hurry never does any good, I have amended. $\endgroup$ Oct 21, 2023 at 19:50
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The method that comes to my mind is similar to this: https://math.stackexchange.com/a/1396968/473276. Count the number of functions $f: [k] \to [n]$ such that $f(i + 1) \ge f(i) + 3$ for all $i$ (these functions correspond exactly to subsets of size $k$ with all elements differing by at least $3$). Observe that $f$ is such a function if and only if the function $g: [k] \to [n - 2k + 2]$ given by $g(i) = f(i) - 2i + 2$ is strictly increasing.

So the total number of such functions is $\binom{n - 2k + 2}k$. (Take this to be $0$ if $n - 2k + 2 < k$).

This obviously generalises to $\binom{n - (d - 1)k + d - 1}k$.

PS: if $k = 1$, the answer should be $n$, right? Your formula gives $\binom{\frac{n - 1}2 + 1}{1} = \frac{n - 1}2 + 1$, which is smaller in general.

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