<s>[INST] In how many ways can $3$ girls and $5$ boys be placed in three distinct cars that each hold at most three children? 
There are $3$ cars of different makes available to transport $3$ girls and $5$ boys on a field trip.  Each car can hold up to $3$ children. Find the number of ways in which they can be accommodated, if the internal arrangement of children inside the car is considered to be immaterial.

I considered three cases:


*

*$3$ G in car 1, $3$ B in car 2, and $2$ B in car 3.This can be done in $\binom{3}{3}\binom{5}{3}\binom{2}{2}3!=60$ ways;

*$2$ G, $1$ B in car 1; $1$ G, $2$ B in car 2; and $2$ B in car 3.  This can be done in $\binom{3}{2}\binom{5}{1}\binom{1}{1}\binom{4}{2}\binom{2}{2}3!=540$ ways;

*$1$ G, $2$ B in car 1; $1$ G, $2$ B in car 2; and $1$ G, $1$ B in car 3.  This can be done in $\binom{3}{1}\binom{5}{2}\binom{2}{1}\binom{3}{2}\binom{1}{1}\binom{1}{1}\frac{3!}{2!}=540$ ways.
The total number of ways thus equals $540+540+60=1140,$ but the answer given is $1680$. Where did I go wrong?
 [/INST] There is no need to distinguish between boys and girls. Since there are $8$ children for nine seats available in the three cars I suggest a $9^{\rm th}$ child that will withdraw at the last moment.
There are ${9\choose3}$ ways to choose the $3$ children for the first car, then ${6\choose3}$ ways to choose the three children for the second car, and the remaining $3$ children will be put in the third car. It follows that there are
$${9\cdot8\cdot7\cdot6\cdot5\cdot 4\over(3!)^2}=1680$$
ways to accommodate the children.
</s>