the answer is no. you might wonder why?
if any of the numbers is divisible by 19, then exactly one is divisible by 19, so we cannot have $\displaystyle \prod_{x\in B} x = \prod_{x\in C} x$.
so none of the numbers is divisible by 19.
However if none of the numbers is divisible by 19, let us denote by $a$ the residue of the product of the whole numbers modulo 19, and $b$, $c$ the residues of $\displaystyle \prod_{x\in B} x,\ \prod_{x\in C} x$ modulo 19 as well.
Then we must have $b=c$, and also $a=bc$, so $a=b^2$, so in fact $a$ is a quadratic residue.
Using Wilson's theorem, we obtain that $18! \equiv -1 \pmod {19}$, but it is well known that if $p\equiv 3 \pmod 4$ then $-1$ cannot be a square residue and we are done.
Indeed this is quite easy.