The answer to your question is affirmative: Yes, there is an intuitive interpretation for the number of choices.
We know at least two interpretations of the binomial coefficient $\binom{n}{k}$ which are useful for combinatorial proofs. These are
Set theoretical view: $\binom{n}{k}$ as number of $k$-element subsets of an $n$-element set
Geometrical view: $\binom{n}{k}$ as number of lattice pathes of length $n$ containing $k$ horizontal $(1,0)$-steps and $n-k$ vertical $(0,1)$-steps
The geometrical view is valid because there are $\binom{n}{k}$ choices to select $k$ steps in horizontal direction leaving the remaining $n-k$ steps for the vertical direction.
Let's apply the second approach to
\begin{align*}
\binom{x}{y}\binom{y}{z}\qquad z \leq y \leq x\tag{1}
\end{align*}
When looking at $\binom{y}{z}$ you may think of all paths of length $y$ starting at $A=(0,0)$ and ending in $B=(z,y-z)$ thereby containing exactly $z$ horizontal $(1,0)$-steps and $y-z$ vertical $(0,1)$-steps. All these paths are enclosed within a rectangle of length $z$ and height $y-z$.
Similarly we consider $\binom{x}{y}$ as the number of lattice paths of length $x$ containing $y$ $(1,0)$-steps and $x-y$ $(0,1)$-steps. But here we let all these paths start in $B=(z,y-z)$ and so they end up in $C=(z+y,x-z)$.
We observe following
Combinatorial interpretation of $$\binom{x}{y}\binom{y}{z}$$
This is the number of lattice paths of length $x+y$ starting in $A=(0,0)$ ending up in $C=(z+y,x-z)$ and passing through the point $B=(z,y-z)$.
The paths contain only horizontal $(1,0)$-steps and vertical $(0,1)$-steps.
