Group of 28 divided into 4 teams, probability of at least 1 girl on each team when there are 7 girls. Hello we had a group divided into 4 teams of 7 each and there is a girl on each team. One of the girls thinks there was foul play in choosing the teams so I thought I'd calculate the chance of a girl on each team. This proved to be harder than I expected so I'm asking for help.
I think that since there are 28 total and 7 girls that we would have $\binom{28}{7}\binom{21}{7}\binom{14}{7}$ ways to assign the people to labelled teams.
I think now I would need to count the number of ways girls could be on teams and then divide?
I am unsure if I am doing anything correctly at this point, been years since I took math.
 A: You only need consider ways to arrange the 7 girls among 28 places (4 teams of 7 places each), and use the principle of inclusion and exclusion to count ways that some teams have no girls.
$$\mathsf P(\text{no team has no girls})=1- \dfrac{{\dbinom 41\dbinom{21}7}-{\dbinom42\dbinom{14}7}+{\dbinom 43\dbinom 77}}{\dbinom{28}7}=\dfrac{16807}{26910}$$

Remark: How does this work?
A method of fairly assigning teams might be to draw names out of a hat, or such lottery, then assign teams based on order of draw.  There are $28!$ ways to do so.  
Now this process is equivalent drawing girls and boys from separate hats, with a fair way to select 7 from 28 positions in the draw to come from the girls' hat. We can see that indeed:$$28!=\binom{28}7 7!\,21!$$
So we just need to focus on the selection of those positions.
Okay, there are $\tbinom 41$ ways to select a team to have all boys and $\tbinom {21}7$ ways to select positions among the rest for the girls.  However, this includes cases where two or more teams have all boys, so we use the principle of exclusion and inclusion to avoid overcounting.  Thus the numerator above.
