Consider a set $X := \{x_1, \ldots, x_n\}$ with corresponding weights $p_1, \ldots, p_n$. Suppose we would like to draw $m < n$ distinct (i.e. unique) elements in a way that the probability of drawing a set $(x_{i_1}, \ldots, x_{i_m})$ is proportional to the product of corresponding weights, i.e. \begin{equation}\tag{1} p_{(i_1, \ldots , i_m)} :\propto p_{i_1} \cdots p_{i_m}. \end{equation} Since we are drawing among all sequences $(i_1, \ldots, i_m)$ with no duplicates, the normalizing constant is the sum over all such sequences: $$\tag{2} p_{(i_1, \ldots , i_m)} = \frac{p_{i_1} \cdots p_{i_m}}{\sum_{\substack{(j_1, \ldots, j_m) \\ \text{no duplicates}}} p_{j_1} \cdots p_{j_m}}. $$ This is equivalent to a sampling scheme where one samples $m$ weighted elements with replacement and then rejects (and repeats) the entire sample if it contains at least one pair duplicate values. Surprisingly (to me) this scheme is not equivalent to weighted sampling without replacement, i.e. where one sequentially samples first $i_1$ with weights, updates all remaining weights and then samples $i_2$, etc.
When $n$ is large, direct sampling by computing the denominator in (2) is practically impossible. Further, if $m$ is close to $n$ or if weights have a very non-uniform distribution, the rejection sampler may take very long before the sampling with replacement manages to draw a sample of $m$ distinct draws. Yet (1) has a simple structure, which perhaps could enable more efficient sampling.
My question is: Is there a way to sample efficiently from this scheme? This could for instance be a sequential sample with some update of weights.