3 different families have 3 children each. In how many ways we can put them into 3 tents, such that in every tent, there are at least 2 siblings? 3 different families have 3 children each. In how many ways we can put them into 3 tents, such that in every tent, there are at least 2 siblings?

Finding the total different ways is easy, just $\binom{9}{3} * \binom{6}{3} * \binom{3}{3}$.
Then I named them, so $ A_1, A_2, A_3, B_1,B_2, B_3, C_1, C_2, C_3 $
I'll fast forward casework, these are the cases we don't want;
$[ABC]-[AAB]-[BCC]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * 3! = 3^4 * 2^2$
$[ABC]-[BBC]-[CAA]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * 3! = 3^4 * 2^2$
$[ABC]-[CCA]-[ABB]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * 3! = 3^4 * 2^2$
$[ABC]-[ABC]-[ABC]$ : $\binom{3}{1} * \binom{3}{1} * \binom{3}{1} * \binom{2}{1} * \binom{2}{1} * \binom{2}{1} = 3^3 * 2^3$
In total, $3 * 3^4 * 2^2 + 3^3 * 2^3 = 3^3 * 2^2 * (3^2 + 2)$
So the probability of not happening is: $\frac{3^3 * 2^2 * (3^2+2)}{\binom{9}{3} * \binom{6}{3}} = \frac{99}{140}$
Finally, $1 - \frac{99}{140} = \frac{41}{140}$,
This was the solution, and the thing is I just hate casework, there is always the feeling of missing something. I was looking for if there was an another elegant solution that uses something/or a technique that are generally applicable, other than casework?

I'm assuming that the tents are indistinguishable and can contain $\leq3$ children.
First put the children family-wise into the  tents, so that we have the three tents
$$(A_1,A_2,A_3),\quad(B_1,B_2,B_3),\quad(C_1,C_2,C_3)\ .$$
No we can make certain permutations, moving $\leq1$ children from each tent:

*

*Leave it as it is. Makes $1$ case.

*Make one transposition. Choose the two involved tents and one child in each of these. These two children will be interchanged. Makes $3\cdot3^2=27$ cases.

*Make a cyclic permutation. Choose one child in each tent and the direction of the circuit. Makes $3^3\cdot2$ cases.

It follows that there are $1+27+54=82$ admissible placements.
