 Solution

For the uniform prior, write $s=x+y$ for the pooled success count in $m+n$ trials. Conditional on $s$, exchangeability makes all placements of the successes equally likely. The successes in the second batch therefore have a hypergeometric distribution, and the factor in question isThis is hypergeometric allocation of exchangeable Bernoulli counts: sample $n$ positions without replacement from a population of $m+n$ positions containing $s$ successes. The uniform-count predictive property gives $\mathrm{Pr}(X+Y=s)=1/(m+n+1)$ and $\mathrm{Pr}(X=x)=1/(m+1)$, soAs $y$ varies with $x$ fixed, $s=x+y$ also varies. Therefore $B$ alone is not a normalized hypergeometric distribution over $y$; its normalization factor is essential.