<s>[INST] In how many ways can $4$ boys and $6$ girls be put into $3$ groups of $3$ people so that there is a boy in each group? In how many ways can $4$ boys and $6$ girls be put into $3$ groups of $3$ people so that there is a boy in each group?
I tried by inclusion exclusion principle where: $$ A_{i} = \left\{\text{there is a boy in }i\text{-th group} \right\} $$ and I would look for the complement.
So my answer would be :
$$ \binom{10}{3}\binom{7}{3}\binom{4}{3} - \binom{3}{1}\binom{6}{3}\binom{7}{3}\binom{4}{3} + \binom{3}{2}\binom{6}{3}\binom{3}{3}\binom{4}{3} - 0$$
 [/INST] I would condition on whether the one left out is a boy or a girl.  If the one left out is a boy, there is one boy in each group.  You can  choose the first group in $4{6 \choose 2}$ ways, the second in $3{4 \choose 2}$ and the third in $2{2 \choose 2}$ ways, giving $$4{6\choose 2}3{4\choose 2}2{2 \choose 2}=2160$$
If a girl is left out there must be two boys in one group and one in each other.  You have $3$ ways to choose the group that gets two boys, $4 \choose 2$ ways to choose the boys, and $6$ ways to choose the girl.  You can then choose the boy for the first remaining group in $2$ ways and the girls in $5 \choose 2$ ways.  For the third group you have one boy and $3 \choose 2$ ways to choose the girls, giving
$$3{4\choose 2}6\cdot 2{5\choose 2}{3 \choose 2}=6480$$
That gives an overall total of $8640$
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