Combinatorics question on group of people making separate groups If there are $9$ people, and $2$ groups get formed, one with $3$ people and one with $6$ people (at random), what is the probability that $2$ people, John and James, will end up in the same group?
I'm not sure how to do this. So far, I've got:
The total number of groups possible is $${9\choose 6}=\frac{9!}{6!3!}=84$$ 
The total number of groups when they are together is $${7\choose 4} +{7\choose 1} =\frac{7!}{4!3!}+\frac{7!}{6!}=42$$
Therefore, probability $= \frac{42}{84} =50\%$
However, I am not sure if that is correct.
 A: Another route: $$\frac39\times\frac28+\frac69\times\frac58=\frac12$$
The first term stands for the probability that both end up in the group with size $3$ and the second term stands for the probability that both end up in the group with size $6$.
Some explanation: there is evidently a probability of $\frac39$ that John ends up in the group with size $3$. Under that condition the probability that James will also end up in that group is $\frac28$ (there are $8$ candidates left for $2$ places).
A: The total number of ways to select the groups is to count how many ways there are to select the first group:
$$
\binom{9}{3} = 84
$$
The number of cases where John and James are together are the cases where they are in the group of 3 (they are fixed, need to select one of the remaining $9 - 2$ as third participant, $\binom{9 - 2}{3 - 2}$ cases) or in the group of 6 (similarly, $\binom{9 - 2}{6 - 2}$ cases). In all:
$$
\binom{9 - 2}{3 - 2} + \binom{9 - 2}{6 - 2} = 42
$$
A: P(both in same group) = P(both in group of 3) + P(both in group of 6) 
$$ =\frac{{3\choose 2}+{6\choose 2}}{9\choose 2} = \frac12$$  
