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

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

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