Let $n,d,\ell\ge1$ be integers, $n\ge 1+(d-1)\ell$. Prove that $$n-(d-1)\ell-1\le\binom{n-(d-1)(\ell-1)-1}{d}$$ and that equality holds if and only if $d=1,\frac{n-1}{\ell}+1,\frac{n-2}{\ell}+1$.
The inequality obviously holds. Moreover, the case $d=1$ is obvious. I only checked numerically that this is true, but I'm having trouble to find a formal proof.
Any help is appreciated.
On
If $n-(d-1)(\ell-1)-1 < d$, then the right hand side is $0$ and the left hand side is $n-(d-1)\ell-1=n-(d-1)(\ell-1)-1-d+1<1 \le 0$.
So, in that case the inequality holds.
Now, assume $n-(d-1)(\ell-1)-1 \ge d$.
The right hand side is the number of ways to choose d objects.
The left hand side is the number of ways to choose $d$ objects such that the first $d-1$ objects in the set are fixed as part of the subset and then we choose one more object to complete the subset from the remaining $n-(d-1)\ell-1$ objects.
Equality can happen only if the two ways of counting give the same number of subsets, so you would need to have $n-(d-1)\ell-1$ equal 0 or 1 -or- you would need $d-1=0$.
John had posted a nice combinatorial proof for it but since I see lalaland's comment, I would like to post a somewhat elementary proof.
The inequality doesn't have to have complicated looking like that, in fact, we can set $x=n-(d-1)\ell-1 \ge 0$ so the inequality becomes: $$x \le \binom{x-1+d}{d} $$ Now if $x=0$ then both sides are $0$ and if $x=1$ then both sides are $1$.
Suppose $x\ge 2$, the inequality has become: $$(x+d-1)(x+d-2) \ldots (x+2)(x+1)(x-1) \ge d!$$ Since $x\ge 2$, $$L.H.S \ge (d+1)d\ldots 4.3.1 \ge d!.$$ Note that in the last inequality, equality holds if and only if $d+1=2$ or $d=1$.