A teacher wants to randomly form two teams of 5 students from a group of 5 girls and 5 boys for a sports activity. Two of the girls, Ann and Alice, are selected as team leaders. Find the probability that one team has exactly 3 girls.
If i do this way $$\dfrac{{3 \choose 2}{5\choose 2}{1\choose 1}{3\choose 3} \times 2!}{{8 \choose 4}{4 \choose 4}} = \dfrac{6}{7}$$
Which turns out to tally with the answer. But i got one big question, is my way correct. Do i really need to times $2$ here? Please see my link below to my previous question cause the previous question make me doubt if i should times $2$.
The team of e.g. Ann goes along with a selection of $4$ persons out $5$ boys and $3$ girls.
So the probability that her team will have exactly $3$ girls is: $$\frac{\binom32\binom52}{\binom84}$$
The same is true for the team of Alice, and the two events are mutually exclusive.
Then the probability that one of the teams has exactly $3$ girls is:$$\mathsf P(\text{team of Ann has }3\text{ girls})+\mathsf P(\text{team of Alice has }3\text{ girls})=2\cdot\frac{\binom32\binom52}{\binom84}=\frac67$$