Each of 10 passengers board any of the three buses randomly which had no passenger initially. The probability that each bus has got at least one passenger is?
Total number of ways $=3^{10}$
Every bus must have atleast one passenger. Let all 3 buses be filled with one passenger each. This can be done in $\binom{10}{3}3!$.
Remaining passenger have 3 choices. So total number of ways is $\binom{10}{3}3!3^7$
$$\text{Probability}=\frac{\binom{10}{3}3!3^7}{3^{10}}$$
But answer given is $$\frac{3^{10}-3\cdot 2^{10}+3}{3^{10}}$$
Best Answer
As @Andre Nicolas suggested, I used principle of inclusion and exclusion.
Total number of ways $=3^{10}$
But I have also counted the number of ways where one bus remains empty. I need to subtract those cases. Number of ways where one bus remains empty $=\binom{3}{1}2^{10}$, where $\binom{3}{1}$ is for choosing the bus.
But I have subtracted the case where 2 buses are empty twice. I need to add that back. Number of ways where 2 buses are empty $=\binom{3}{2}1^{10}$
$$\text{probability}=\frac{3^{10}-\binom{3}{1}2^{10}+\binom{3}{2}1^{10}}{3^{10}}$$
The mistake I made was pointed out by @Andre Nicolas in a comment to the original post: