Gap in binomial coefficients

We indeed have $z = 0$. I don't think this at all answers the question "What is going on with this figure?".

Take $\varepsilon > 0$. We show the limsup is at most $\varepsilon$. Take large $C = C(\varepsilon) \in \mathbb{N}$. By a "union-bound", the number of pairs $a,b \le x$ with $\gcd(a,b) > C$ is at most $\sum_{g > C} \lfloor \frac{x}{C} \rfloor^2 \le \frac{10x^2}{C} \le \varepsilon x^2$.

So let's restrict to pairs $(a,b)$ with $\gcd(a,b) \le C$. This means there are only finitely many possible prime powers that divide both $a$ and $b$. For example, if $C = 10$, then the possibilities are $2^1, 2^2, 2^3, 3^1, 3^2, 5^1, 7^1$. Let $K \in \mathbb{N}$ be the largest exponent among these possible prime powers. Note that there are at most $C$ possible primes composing these prime powers, and note that $K \approx \log_2 C$.

Now, $a!b! \mid (a+b-1)!$ iff for each prime $p$, we have $$\sum_{k=1}^\infty \left(\left\lfloor \frac{a+b-1}{p^k} \right\rfloor - \left\lfloor \frac{a}{p^k} \right\rfloor - \left\lfloor \frac{b}{p^k} \right\rfloor\right) \ge 0.$$ The key (and trivial) observation is that \begin{equation}\tag{1}\label{1} \left\lfloor \frac{a+b-1}{p^k} \right\rfloor - \left\lfloor \frac{a}{p^k} \right\rfloor - \left\lfloor \frac{b}{p^k} \right\rfloor = \begin{cases} -1 & \text{if } r_a, r_b = 0 \\ 0 & \text{if } 0 < r_a + r_b < p^k \\ 1 & \text{if } r_a+r_b \ge p^k \end{cases}, \end{equation} where $r_a, r_b$ are the remainders when $a$ and $b$ are divided by $p^k$, respectively.

Let's focus on equation \eqref{1} for $k \le 100K^2$, say. For $k > 100K^2$, we don't get $-1$ (by the definition of $K$).

Since $100K^2$ is some fixed number, it holds as $x \to \infty$ that if we choose $(a,b) \in [x] \times [x]$ uniformly at random subject to a $\gcd$ constraint, then the residues $(r_a,r_b)$ are equidistributed (up to error $o_{x \to \infty}(1)$) in a suitable sense. A bit more precisely, once we condition on $\gcd(a,b) = g$ with $p^{k_0} \parallel g$, the residues $(r_a,r_b)$ modulo $p^{k_0+1}$ are equidistributed over the set $\{0, p^{k_0}, 2p^{k_0},\dots, (p-1)p^{k_0}\}$, and once we condition on the residues modulo $p^{k_0+1}$, the residues modulo $p^{k_0+2}$ are distributed uniformly over an analogous set. Etc. up to $p^{100K^2}$. Therefore, whether the sum of the residues is at least the modulus (i) has probability at least $\frac{1}{10}$, say and (ii) is independent of the previous residues.

[Remark: about (i), the probability is close to $1/2$ as $p$ increases, but can be smaller, for example if $p^{k_0} = 2$, then the residues modulo $p^{k_0+1} = 4$ are $\{0,2\}$, giving a probability of $\frac{1}{4}$].

Consequently, although in the worst case we get a sum of $-K$ for $k \in \{1,\dots,K\}$, we, with probability $1 - \exp(-K^2)$ (over the choice of $(a,b)$), end up with a non-negative (in fact, positive and large) sum for $k \in \{1,\dots,K^2\}$.

By union-bounding over all possible primes, of which there are at most $2^K$, we get probability at least $1 - o_K(1) \ge 1 - \varepsilon$, for $a!b! | (a+b-1)!$, as desired.