# 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 : $$\binom{10}{3}\binom{7}{3}\binom{4}{3} - \binom{3}{1}\binom{6}{3}\binom{7}{3}\binom{4}{3} + \binom{3}{2}\binom{6}{3}\binom{3}{3}\binom{4}{3} - 0$$

• are the groups distinguishable? – drhab Mar 10 at 12:24
• It's not specified but let's say they are – user15269 Mar 10 at 12:28
• For certainty: $3$ groups of $3$ gives place for $9$ children. Is one of the $4+6=10$ left out? – drhab Mar 10 at 12:31
• Yes, one of them is left out – user15269 Mar 10 at 12:34
• Can you compute the probability in a simpler case, where there is only 1 group with 1 boy and 2 girls? – Ertxiem - reinstate Monica Mar 10 at 13:05

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 $$8640$$

I preassume that there are $$3$$ distinguishable groups of $$3$$ and label the groups with A,B,C.

If a boy is left out in the sense that he is not placed in one of the groups A,B,C then there are $$\frac{4!}{1!1!1!1!}=24$$ possibilities for placing the boys and $$\frac{6!}{2!2!2!0!}=90$$ for placing the girls.

If a girl is left out in the sense that she is not placed in one of the groups A,B,C then there are $$\frac{4!}{2!1!1!0!}=12$$ possibilities for placing the boys and $$\frac{6!}{1!2!2!1!}=180$$ for placing the girls in such a way that exactly one of the groups A,B,C contains $$2$$ boys.

So finally we find $$24\times 90+3\times12\times180=8640$$ possibilities.

If the groups A,B,C are not distinguishable then above we are dealing with multiple counting that can be repaired by dividing with $$3!=6$$ (i.e. the number of ways the groups can be ordered) and arrive at $$1440$$ possibilities.