Question on union of particular subsets of a given set Given positive integers $n,d,\ell$, $d\ge2$, $n\ge 1+(d-1)\ell$. Let $s$ be the minimum integer possible such that the union of any $s$ subsets $A_1,\dots,A_s$ of the set $\{1,\dots,n\}$ having cardinality $d$ and with the property that for each $i$, $A_i$ consist of elements $a_{i,1},\dots,a_{i,d}$ such that $a_{i,j+1}-a_{i,j}\ge\ell$, for all $j=1,\dots,d-1$, is $A_1\cup\dots\cup A_s=\{1,\dots,n\}$.
Prove that $s$ is the integer
$$
s=\binom{n-(d-1)(\ell-1)-1}{d}+1.
$$
I could only prove the case $\ell=1$. In such case $s=\binom{n-1}{d}+1$ and the subsets $A_i$ are ordinary subsets. That $s=\binom{n-1}{d}+1$, if $\ell=1$ was proved in a previous question, I can add the link if anyone wants to see that. Maybe induction could work?
Any help is appreciated.
 A: Hmm. Say $n=5$, $d=\ell=2$. Then $1+(d-1)\ell = 3$, so $n \geq 1 + (d-1)\ell$ is satisfied. Let
$$
  s = \binom{n-(d-1)(\ell-1)-1}{d} + 1 = \binom{5-1-1}{2}+1 = \binom{3}{2}+1 = 4.
$$
Now consider the sets
$$
  A_1=\{1,4\}, \quad A_2 = \{1,5\}, \quad A_3 = \{2,4\}, \quad A_4 = \{2,5\}.
$$
They are distinct sets, they each have cardinality $d=2$, in each of them the elements are separated by at least $\ell=2$, they are subsets of $\{1,\dotsc,5\}$.
But their union is only $\{1,2,4,5\}$, not all of $\{1,2,3,4,5\}$.
This example shows that this value of $s$ is not large enough.
It's not hard to show that the correct value of $s$ must be at least the value you said.
There are $\binom{(n-1)-(d-1)(\ell-1)}{d}$ subsets of $\{1,\dotsc,n-1\}$ with $d$ elements that are separated by at least $\ell$–use subtraction of multiples of $\ell-1$ to get a bijection with $d$-element subsets of $\{1,\dotsc,(n-1)-(d-1)(\ell-1)\}$. So you can choose $\binom{(n-1)-(d-1)(\ell-1)}{d}$ subsets whose union avoids $n$. Therefore $s > \binom{(n-1)-(d-1)(\ell-1)}{d}$.
But it seems tricky to count the number of $d$-element, $\ell$-separated subsets of $\{1,\dotsc,k-1,k+1,\dotsc,n\}$.
So I don't know what is the true value of $s$, but it can be larger than $\binom{(n-1)-(d-1)(\ell-1)}{d}+1$, I think.
