A group of 9 people was further divided into 3 teams of three persons. How do you find the probability that 2 persons (let's name them person A and person B) are in the same group?
I think this is similar to the word permutation problem wherein two letters must be right next to each other, so you just treat them as one.
What I would do is treat person A and person B as one person and continue in grouping them, but there is a special case that only in the group of this new person can there be only 2 persons and not 3 (since that new person represents two).
Using the combination formula: $\frac{n!}{r!(n-r)!}$
\begin{align*}
&= \dfrac{9!}{3!(9-3)!}\\
&= \dfrac{9!}{3!\cdot6!}\\
&= 84
\end{align*}
This is clearly wrong since it is not taking into account the fact that I am treating two persons as one.
Best Answer
If $A$ gets into any of the teams, there are still two places in the same team.
The probability of $B$ getting into one of those two places out of remaining $8$ people $ = \displaystyle \frac{2}{8} = \frac{1}{4}$.
Or you can think of it as below -
$A$ and $B$ can be together in any of the $3$ teams and whichever team they are on, there is just one slot in the team where $1$ from rest of the $7$ people can get in.
So, desired probability $ \displaystyle = \frac{{3 \choose 1} \times {2 \choose 2 } \times {7 \choose 1} }{9 \choose 3} = \frac{1}{4}$