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.

P(both in same group) = P(both in group of 3) + P(both in group of 6) 
$$ =\frac{{3\choose 2}+{6\choose 2}}{9\choose 2} = \frac12$$  
