Questions concerning Dividing Groups using Combinations 15 people with 10 males and 5 females are divided into 3 groups each of 5. 3 groups are named A, B, C.
Find the number of ways can be grouped if
a) no restriction in grouping 
b) each group has only 1 gender
c) Male A and Male B in the same group
d) Female A and Female B not in the same group
e) Male C, Male D and Female C in different groups
For a), $$
\frac{\binom{15}{5} \binom{10}{5} \binom{5}{5}}{3!}.
$$
is this method correct?
For b), $$
\frac{\binom{5}{5} \binom{10}{5} \binom{10}{5}}{3!}.
$$
is this method correct?
For c), $$
\frac{\binom{14}{5} \binom{9}{5} \binom{5}{5} \binom{2}{1}}{3!}.
$$
is this method correct?
I have no idea about d) and e), please help.
 A: I think there is a misinterpretation of the question. At the beginning, it asks for division into groups, to be named A, B, and C. To put it more colourfully, we are dividing into three groups, to wear uniforms coloured Green, Red, and Yellow. I have no idea why people are also sometimes named A, B, and C. We give details of the analysis, so that if the problem has been mistranslated, you can at least use the ideas.  
In a), if we are using labelled teams, there should not be the division by $3!$. But if we are using unlabelled teams, the division by $3!$ is correct.  
For b), under this labelling interpretation, the answer is $3\binom{10}{5}$. For there are $3$ choices of colour for the female group. For each of these choices, there are $\binom{10}{5}$ ways to choose the males that will wear the alphabetically higher ranking remaining colour. 
Under your interpretation, which I believe is not the intended one (no labels), we are just dividing the $10$ males into two teams. That can be done in $\frac{1}{2}\binom{10}{5}$ ways. 
The answer proposed in the post is not correct under either interpretation.  
c) We go back to the labelled interpretation. There are $3$ choices of colour for the team that contains people A and B. There are $\binom{13}{3}$ ways to select the rest of the team. And now we are dividing $10$ people into two coloured teams, which can be done in $\binom{10}{5}$ ways.
Under the no labels interpretation, there are $\binom{13}{3}$ ways to choose the rest of the team that contains A and B, and then $\frac{1}{2}\binom{10}{5}$ ways to do the rest of the splitting. 
d) For female A and female B not in the same group, use the reasoning of c) to count the number of ways to have female A and female B in the same group, and subtract from the total number of ways to distribute. This works for the labels interpretation just as well as for no labels. We just have to be consistent.
Or else we can attack the problem directly. For labelled teams, for example, there are $3$ ways to choose the colour of female A's group, and for each choice there are $2$ ways to choose the colour of female B's group. Now we choose the rest of female A's group, and the rest of female B's group. 
e) Again, we have a choice between labels and no labels. For a change we will only do the no labels interpretation (so the interpretation that I think incorrect).  
There seem to be $3$ people, two male and one female, who should be in distinct groups. We do the no labels calculation. For labels, multiply by $3!$. 
There are $\binom{12}{4}$ ways to choose the people to join male C's team. For each of these, there are $\binom{8}{4}$ ways to choose the people to join male D's team. And now we are done, though of course we can multiply by $\binom{4}{4}$. Note that for unlabeled teams, we do not divide by $3!$, for there has been no double counting. 
