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You have equality, yes.

The same proof I gave here applies equally well to this situation.

Here's a generalization of both (also implied by that proof).

Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where any $r$ edges share at most $1$ point and each point is contained in at least $r$ edges. Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.

(The modification of the proof is to consider $r$-element subsets of $V_i$.)

You have equality, yes.

The same proof I gave here applies equally well to this situation.

Here's a generalization of both (also implied by that proof).

Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where any $r$ edges share at most $1$ point and each point is contained in at least $r$ edges. Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.

(The modification of the proof is to consider $r$-element subsets of $V_i$.)

You have equality, yes.

The same proof I gave here applies equally well to this situation.

Here's a generalization of both (also implied by that proof).

Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where each edge has size at least $r$, any $r$ edges share at most $1$ point, and each point is contained in at least $r$ edges. Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.

(The modification of the proof is to consider $r$-element subsets of $V_i$.)

You have equality, yes.

The same proof I gave here applies equally well to this situation.

Here's a generalization of both (also implied by that proof).

Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where any $r$ edges share at most $1$ point and each point is contained in at least $r$ edges. Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.

(The modification of the proof is to consider $r$-element subsets of $V_i$.)

You have equality, yes.

The same proof I gave here applies equally well to this situation.

You have equality, yes.

The same proof I gave here applies equally well to this situation.

Here's a generalization of both (also implied by that proof).

Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where each edge has size at least $r$, any $r$ edges share at most $1$ point, and each point is contained in at least $r$ edges. Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.

(The modification of the proof is to consider $r$-element subsets of $V_i$.)