$27$ balls into $3$ cells Spreading $15$ white balls and $12$ black balls into $3$ cells, each of which can contain any number of balls.
$(A.)$ Find the probability that in each cell there will be exactly $5$ white balls.
$(B.)$ Find the probability that in each cell there will be exactly $5$ white balls and exactly $4$ black balls.
$(C.)$ Find the probability that in one cell there will be the whole the black balls and $3$ white balls, and there will be  an even number of white balls in the two other cells.

My attempt:
$(A.)$ $\Omega$ is all the possible spredings $|\Omega|=3^{15}3^{12}=3^{27}$ because each ball "choose a cell"
Let A be the event that in each cell there will be exactly $5$ white balls,
$$|A|=\binom{15}{5}\binom{10}{5}\binom{5}{5}\\
\Longrightarrow P[A]=\frac{\binom{15}{5}\binom{10}{5}\binom{5}{5}}{3^{27}}=\frac{6250}{2541865828329}$$
$(B.)$ Let $B$ be the event that in each cell there will be exactly $4$ black balls. $$|B|=\binom{12}{4}\binom{8}{4}\binom{4}{4}\\
\Longrightarrow A \text{ and } B \text{ are independent I think so: }P[A\cap B]=P[A]P[B]=\frac{\binom{15}{5}\binom{10}{5}\binom{5}{5}}{3^{27}}\cdot \frac{\binom{12}{4}\binom{8}{4}\binom{4}{4}}{3^{27}}=\frac{6250}{2541865828329} \cdot \frac{3850}{847288609443}$$
$(C.)$ Let $C_1$ be the event that in one cell there will be the whole the black balls and $3$ white balls, $|C_1|=\binom{12}{12}\binom{15}{3}\times 3$, and let $C_2$ be the event that there will be  an even number of white balls in the two other cells, $|C_2|=\binom{8}{4}\binom{4}{4}\times 2$.
$C_1$ and $C_2$ are independent so $$P[C_1\cap C_2]=P[C_1]P[C_2]\\
P[C_1]=\frac{\binom{12}{12}\binom{15}{3}\times 3}{3^{27}}\\
P[C_2]=\frac{\binom{8}{4}\binom{4}{4}\times 2}{3^{27}}\\
\Longrightarrow =\frac{910}{7625597484987}\cdot \frac{70}{2541865828329}$$
My attempt is correct?

$(A)$ You distributed all of the balls in the denominator but only distributed the white balls in the numerator. $|A|$ should be
$$
\binom{15}{5}\binom{10}{5}\binom{5}{5}3^{12}
$$
$(B)$ There is a similar error.
$(C)$ If you take away $3$ white balls, $12$ remain not $8$.
