You can use inclusion-exclusion when the problem with a more general but opposite constraint is easier. Here you've already computed the solution without constraints. If the constraint would be that the first kid gets at least $6$ oranges, then start giving those oranges, and find $\binom{30-6+9}{30-6}=\binom{33}9$ solutions. But not only could it be another kid that gets too many oranges, there could be more than one kid at once that gets too much. So the "more general opposite" constraint would be, for any set $S$ of kids, that all kids from $S$ get at least $6$ oranges. For this there are $\binom{30-6s+9}{30-6s}=\binom{39-6s}9$ solutions if $S$ has $s$ kids in it.
By inclusion-exclusion you need to count the solutions for $S=\emptyset$, subtract those for $S$ a singleton, add back for $S$ a doubleton, etc. All in all you get $ \binom{39}9-\binom{10}1\binom{33}9+\binom{10}2\binom{27}9 -\binom{10}3\binom{21}9+\binom{10}4\binom{15}9 -\binom{10}5\binom99 $ solutions, which if I calculated well gives $2\,930\,456$ possibilities, less than $1.5$% of the original $211\,915\,132$.
Added (much later). Alternatively, one could compute the coefficient of $X^{30}$ in $(1+X+X^2+X^3+X^4+X^5)^{10}=\frac{(1-X^6)^{10}}{(1-X)^{10}}$. This is quite easy (the numerator has only $6$ terms of degree${}\leq30$) and gives the same result, in fact even via the same formula.