Splitting white and black balls into bowls of three

There are $7$ identical white balls and $8$ identical black balls, and $5$ distinguishable bowls. In how many ways can all the balls be placed so that every bowl ends up with exactly $3$ balls?

$155$. Once each bowl's number of white balls is set, black balls fill the rest uniquely, so count $w_1+\dots+w_5=7$ with $0\le w_i\le3$. Stars and bars gives $\binom{11}{4}=330$; subtract the $5\binom74=175$ solutions where some bowl has at least 4 (two such bowls is impossible), leaving $155$.