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!
Best Answer
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 $x_1=x_2=x_3=x_4=3$.