Combinatorial meaning of a reduced fraction in a simple probability problem?

Question

A routine exercise for undergraduates says: Given that the number of successes in $20$ independent Bernoulli trials was $8,$ what is the conditional probability that exactly $3$ of those $8$ successes were in the first $10$ trials?

The answer (which actually does not depend on the probability of success on each trial, that information having been superseded by the event that there were $8$ among $20$) comes to $$ \frac{\binom{10}3 \binom{10}5}{\binom{20}8} \approx 0.240057156 $$ A fact treated as something of no interest in such exercises is this: \begin{align} & \frac{\binom{10}3 \binom{10}5}{\binom{20}8} = \frac{30\,240}{125\,970} = \frac{30\times1008}{30\times4199} \\[8pt] = {} & \frac{30\times(2^4\times3^2\times7)}{30\times(13\times17\times19)} \approx 0.240057156 \tag1 \end{align}

My question is whether the integers in the line $(1)$ above, and in particular those within the parentheses, have some interesting combinatorial significance?

Answer 1

It is more or less equivalent to ask about the $p$-adic valuation of the answer, for all primes $p$. Here is is a standard fact that $$ v_p(n!) = \sum_{k>0} \lfloor n/p^k\rfloor = (n-\alpha_p(n))/(p-1), $$ where $\alpha_p(n)$ is the sum of the base $p$ digits of $n$. This means that $$ v_p\binom{i+j}{i} = \frac{\alpha_p(i)+\alpha_p(j)-\alpha_p(i+j)}{p-1}, $$ which is a measure of how much carrying takes place when we add $i$ and $j$ in base $p$. There are various applications of this kind of thing in algebraic topology, via the structure theory of modules over the Steenrod algebra.