Having difficult time understanding the solution to this probability problem So, the task says: there are 12 passengers and 4 wagons, what is probability that 3 passengers entered every wagon?
this is the answer $\frac{\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}}{4^{12}}$ but why instead of $4^{12}$ it doesn't go $12^{4}$ ? I understand that for first wagon we chose 3 out of 12 passenger and hence $\binom{12}{3}$ and so on, but should the same logic be applied for the denominator, shouldn't it be, for the first vagon we can chose 12 passener, for the second the same, and so on?
Thank you!
 A: Each passenger has $4$ options, therefore the number of distributions is $4^{\text{number of passengeres}}$. Note that is such distributions we can have two different passengers on the same wagon. On the other hand, $12^4$ means that each wagon has $12$ options and the same passeger could be assigned to $2$ different wagons. This is impossible!
Moreover note that
$$4^{12}=(1+1+1+1)^{12}=\sum_{x_1+x_2+x_3+x_4=12\\x_1\geq 0,x_2\geq 0,x_3\geq 0,x_4\geq 0}\frac{12!}{x_1!x_2!x_3!x_4!}
\\=\sum_{x_1+x_2+x_3+x_4=12}\binom{12}{x_1}\binom{12-x_1}{x_2}\binom{12-x_1-x_2}{x_3}\binom{12-x_1-x_2-x_3}{x_4}.$$
You are interested in the distribution where $x_1=x_2=x_3=x_4=3$.
A: There is not actually formal consensus over your question.For example @Brian M. Scott accept both of them as true , as far as i remember from earlier questions.
However, if i were you , i would think this types of combinatorics problems like real world problems . What i meant was that who has decision rights for choosing ? Wagons or people ?
The answer is obviously people , so let people to choose which wagon they want to get in. First people has $12$ choices , second has $12$ choices etc..
You can extend this question for letters and postboxes. You put the letters into post boxes in real world.So , if there are $3$ letter and $5$ post boxes ,you have $5^3$ different choices to put the letters into post boxes.
A: Let's say the $12$ passengers are $A, B, C, D, E, F, G, H, I, J, K, \text{and} 
\space L$. We will follow your reasoning and see the loophole. Wagon 1 has 12 options to pick a passenger, agreed. Let's say it picks $A$. You said Wagon 2 has $12$ options as well, here's where the problem is: those $12$ options include Wagon 2 picking $A$ as well. However, $A$ cannot go into two different wagons at the same time!

We do $4^{12}$ instead because it does the reverse: counting how many ways there are for each passenger, not each wagon. There are $4$ wagons, so $4$ ways for $A$, $4$ for $B$, etc.
This doesn't have the same problem as $12^4$ because one wagon can have more than one person. If it were a constraint that a wagon can have at most $1$ person, then this way wouldn't work either.

I hope this helps; feel free to ask about any unclearness in this post.
