If $12$ distinct balls are to be placed in $3$ identical boxes, then what's the probability that one of the boxes contains exactly $3$, balls. This is a bonus question asked in a competitive examination. It all choice are wrong.
I will put the question, then the solution I got from the internet and then my solution.
Please help to figure out whether my solution is correct or not.

The Answer of the website is as follow and I presume that it is wrong

My solution is as follow

Please help me with finding the correct solution if my solution is wrong.
 A: Whether the urns are identical or distinct does not change the probability that one of the boxes has exactly $3$ balls.
Applying the principle of inclusion exclusion, number of ways of distributing $12$ balls in three urns such that at least one of the urns has exactly $3$ balls is
$\displaystyle {3 \choose 1} {12 \choose 3} \cdot 2^9 - {3 \choose 2} {12 \choose 3} {9 \choose 3}$
We first choose an urn that will receive $3$ balls and choose $3$ balls for the urn and then distribute rest $9$ balls in remaining $2$ urns. But each of them will also count cases where two of the urns have $3$ balls each. So in total, cases where two urns receive $3$ balls each is overcounted once and hence we subtract those using the next term.
The question does not state clearly but if the distribution must have balls in each of the three urns then the number of ways is,
$\displaystyle {3 \choose 1} {12 \choose 3} \cdot (2^9 - 2) - {3 \choose 2} {12 \choose 3} {9 \choose 3}$
I take out two cases from $2^9$ where one of the remaining urns received remaining all $9$ balls.
Finally, dividing by $3^{12}$ gives the desired probability.
A: Yes, all the answers are wrong.
Your mistake lies towards the final step.
Distributions which have exactly one box having exactly $3$ balls is $29,920$ and if this is the intent, this should be the numerator.
Else if $3-3-9$ cases are also to be considered, you need to add only half of $18,480$
Thus final answer should either be $\dfrac{29920}{88574}\;\; or\;\;  \dfrac{39160}{88574}$
Added material for $3$ ball cases
$\displaylines{9-3-0:\frac{12!}{9!3!} = 220\\ 8-3-1:\frac{12!}{8!3!1!}=1980\\6-3-3: \frac{12!}{6!3!3!}\Big/2! = 9240\\5-4-3: \frac{12!}{5!4!3!} = 27720}$
Total if exactly one box has to have exactly $3$ balls $= 29,920$
Total if one or more boxes can have exactly $3$ balls $=39,160$
A: The probability is equivalent to the sum of the probabilites that any urn gets 3 balls, that is, "3 balls go in urn 1 and 9 go in the other two" + "3 balls go in urn 2 and 9 go in the other two" + "3 balls get in urn 3 and 9 go in the other two".
You can compute these probabilites by just assigning urn numbers to each ball one by one. You'd find that the probability is 1/3^3 * 2/3^9 * 12 C 3, multiplied by 3 for each of the urn gives an exact 112640/177147, which is the same result as given by the website
