Six children are playing dodgeball. Each child has a ball. At the sound of the whistle, each child chooses another child at random to chase. What is the probability that there is at least one pair of children who are chasing each other?
This is a pretty common type of problem, and usually, the solution would go like this, where we use PIE:
- For one or more pairs, there are $\binom{6}{2}\cdot 3^4=1215$ choices.
- For two or more pairs, there are $\frac{\binom{6}{2} \binom{4}{2}}{2}=45$ choices.
- For three or more pairs, there are $\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$ choices.
Then, we do $1215-45+15=1185$ for undercounting and overcounting, and the answer is $1185/5^6$=$237/3125$. However, for the second bullet point, we are choosing three pairs even though we are only explicitly choosing two pairs, because the third one is predetermined after we choose $2$. Is this solution correct, or is the third term redundant, and the second term should be replaced with $\frac{\binom{6}{2}\binom{4}{2}}{3!}$?