Problem Statement:
We have $n$ distinct positive integers say $a_1,a_2....a_n$ and a given integer value $M$. We have to count number of ways to distribute these integers to $r$ identical bins subject to the constraints that the product of integers in any bin can not exceed $M$ and also no bin should be empty and we must distribute all the integers to some bin or the other.
My thoughts:
I believe this is a type of distribution of distinct objects into identical bins but with a constraint on each bin. The constraint function here is the product of integers in the bin, but it can be sum of integers in a bin or number of integers in a bin or anything else. Like say this or this the problem's limiting function is number of balls in a bin. But unlike our case in both of them balls are identical
Approach1 on top of my mind:
As point out by @Sil If the problem was about limit on sum the following approach would have worked though not very efficient.
Lets say my integers are $2,5,7,11$.
If $r=1$ then its equivalent to counting sum of all coefficients of $x^i$ for $1 \le i\le M$ in $(1+x^2)(1+x^5)(1+x^7)(1+x^{11})$
Similarly if $r=2$ then its equivalent to counting sum of all coefficients of $x_1^ix_2^j$ for $1 \le i,j\le M$ and in $(1+x_1^2+x_2^2)(1+x_1^5+x_2^5)(1+x_1^7+x_2^7)(1+x_1^{11}+x_2^{11})$
......
......
Similarly if $r=r$ then its equivalent to counting sum of all coefficients of $x_1^ix_2^j..x_r^w$ for $1 \le i,j,..w\le M$ and in $(1+x_1^2+x_2^2+..+x_r^2)(1+x_1^5+x_2^5+..+x_r^5)(1+x_1^7+x_2^7+..+x_r^7)(1+x_1^{11}+x_2^{11}+..+x_r^{11})$
This is $O(r^n)$, not very efficient to say the least, so there surely exists more efficient ways to get the result.
Another way of thinking is:
I think it can be solved by PIE let's call the resut $f(r$, set of all bins$)$, we count unrestricted, then fix a particular bins split its restrcted and unrestricted case an then call $f(r-1$, remaining $r-1$ bins$)$ , now fix two boxed and add back the same $f(r-2$, remaining $r-2$ bins$)$ and so on. My idea are still vauge, needs little bit more thought.
Another problem:
I have another part of this problem, let me know if solving this is easier. Lets say we have the same numbers $a_1,a_2....a_n$ and the same $M$. Now if I say we dont care about number of bins you use. Form new set of numbers fusing together(i.e product I mean) two or more of them such that no such fused number is greater than M. How many such sets can you form. Here we must use all the numbers.
I was initially thinking about solving the first part and then sum for each r=1 to r for this part but now I feel it might be easier to approach this directly.
Example clarifying this part: intial_nos = {2,5,7,11}, M =70
case 1:{2,5,7,11}
case 2:{10,7,11}
case 3:{5,14,11}
case 4:{5,7,22}
case 5:{2,11,35}
case 6:{2,7,55}
case 7:{70,11}
So answer should be 7