Let $X, Y$ be positive real numbers, and let $p\_1, ... p\_n$ be the primes less than $Y$. How many subsets $S$ of the integers from 1 to $n$ are there such that the product of the $p\_i$'s with $i$ in $S$ is less than $X$, as a function of $X,Y$?


Clearly, if $X$ is sufficiently large relative to $Y$, then all choices of $S$ work. I am interested in the case when $Y$ is not too small compared to $X$, say $Y = X/C$ for some fixed positive constant $C$ larger than 1.



Number of products of distinct primes lying in an interval
Distinct sets of primes have distinct products, hence you are counting the number of square-free $Y$-smooth integers below $X$.


The number of $Y$-smooth integers below $X$ is
$$\Psi(X,Y)=X\rho(u)+O(X/\log Y),$$
where $X=Y^u$, and $\rho$ is Dickman’s function, which is $1-\log u$ for $u\in[1,2]$. In the range you are interested in (or more generally, if $X=Y^{1+o(1)}$), we have $u=1+o(1)$, hence $\rho(u)=1+o(1)$, and
$$\Psi(X,Y)=X+o(X).$$
In other words, the number of integers below $X$ that are *not* $Y$-smooth is $o(X)$. Since there are $(6/\pi^2)X+O(\sqrt X)$ square-free integers below $X$, the number of your sets is
$$\frac6{\pi^2}X+o(X).$$


