How to count fixed-sized subsets of pairwise co-prime numbers less than a prime, satisfying an additional ‎constraint‎? In part of my research, I need to count (or find a polynomial bound for) the number of ‎possible ‎ways to select $n$ distinct integers less than the prime $p$, say $r_1, r_2, …, r_n$, which ‎are pairwise co-prime and the product of ‎any $k$ of them ($k<n$) is less than $p$. I call the ‎number I look for $\alpha$.‎
Let $p_i$ denotes the $i$-th prime number, and $P=\{ p_1, p_2, …, p_s\}$ denotes the set of all primes less than ‎‎$t=\dfrac{p}{p_{n-k}\cdot p_{n-k+1}\cdots p_{n-2}}$. Having in mind the idea of this answer, I think of these two ‎approaches:‎
‎Approach 1. The number of ways to partition a set of $a$ labelled objects into $n$ ‎nonempty ‎unlabelled subsets, for any $n\leq a \leq s$, is a lower bound for $\alpha$, i.e.‎
‎$$\sum_{a=n}^{s}{s \choose a}S(a,n) \leq \alpha,$$‎
‎where $S(a,n)$ is the Stirling number of the second type.‎
This bound, just, does not count the possibilities that in some of the partitions above, some $p_i$'s ‎can be replaced by ‎$p_i^j$, for some $1 ‎‎< j< \log_{p_i}t$.‎
‎Approach 2. The number of possible selections is bounded by:‎
‎$$ \alpha \leq {{t}\choose{n}}.$$
I am looking forward any idea to improve these bounds.‎
 A: Here is another approach which may give some light on estimating your desired quantity (the number of sets of $n$ numbers $r_i$ which are mutually coprime and whose product is less than $p$). I don't see the significance of $p$ being prime, and will replace $p$ with $m^n$.
Let $R$ be a finite subset of the primes, let $P$ be the product of the primes in $R$ and let $q_m(R)$ count those multiples of $P$ at most $m$ which are $R$-smooth.  Let $S$ be a partition of $n$ parts of some finite subset of primes, with parts called $R_i$.  The number of ways of choosing a number from each part is at least $\prod_i q_m(R_i)$ to get a set of $n$ coprime numbers whose product is at most $m^n$.  This can be tweaked to change the $m$ to $m_i$ whose product is at most $m^n$, but then one does some double counting if insufficient care is taken.
Now a lower bound for your count can be written  using the notation above (where the sum is over all $S$ which are $n$-part partitions of a subset of enough primes) as $$\sum_S \prod_i q_m(R_i).$$  Note that multiple choices of $r_i$ can result in the same product, so it is not enough to consider smooth numbers less than $m^n$.
While the above may help derive a lower bound, it may be more fruitful to consider for each integer $p\lt m^n$ the number of $n$- part coprime factorizations of $p$, then sum that number from $p=1$ to $m^n$.  If I get a good idea for estimation with either scheme, I will add it below.
Gerhard "Happy Father's Day To You" Paseman, 2016.06.19.
