Counting Boys and Girls in a Team There are 12 children and are being divided into two teams to play a game.
$(a)$ In how many ways can you divide 12 children, 6 boys and 6 girls, into two teams to play a game, if the teams are called the Bees and the Sees?
$(b)$ In how many ways can you just divide the children into two teams?
$(c)$ In how many ways can you divide the children into two teams if
each team has to have three boys and three girls?
$Attempt:$
$a)$ $12!/$$2!2!$
$b)$ $12!/$$2!$
$c)$ $3!3!/$$2!$
Is this correct thinking?
 A: Problem a): We have to choose $6$ people from $12$, to be called the Bees. This can be done in $\dbinom{12}{6}$ ways.
Problem b):  Note that we can interchange the names of the teams, and still get the same division into teams. So for b) you need to divide the answer of a) by $2$. 
Another somewhat different way to do b) is to imagine that Alicia is one of the $12$ people. Then the number of ways to divide the group into two nameless teams is the number of ways to choose $5$ people from $11$ to join Alicia. There are $\dbinom{11}{5}$ ways to do this. Of course this is the same as the $\dfrac{1}{2}\dbinom{12}{6}$  
Problem c): Let's solve first a different problem, how can we do it with the teams to be called the Bees and Cees? The $3$ girls for the Bees can be chosen in $\dbinom{6}{3}$ ways, and for each such choice the boys can be chosen in $\dbinom{6}{3}$ ways, for a total of $\dbinom{6}{3}\dbinom{6}{3}$.  Now the Cees are also determined.
But we are dividing into nameless teams, so as in b) we have to divide by $2$.
Or else we could use the "Alicia" argument of b). There are $\dbinom{5}{2}$ ways to choose $2$ girls to join Alicia, and then $\dbinom{6}{3}$ ways to choose the boys.
