Q:

How many ways to choose a group of 7 out of 16 men and 12 women if sex should be proportionally represented

(a) If we have no restrictions, its $\dbinom{28}{7}$
(b) If we need to have each sex proportionally represented: 
Is it possibly so that the answer is $\dbinom{12}{4}$$\dbinom{12}{3}$ 
(c) If we need to have at least on of each sex
$\dbinom{16}{1}$$\dbinom{12}{1}$ $\dbinom{26}{5}$
(d) If the group should be either only men or only women
Here I am thinking that $\dbinom{16}{7}$$\dbinom{12}{7}$ 

A:

Your answer to (d) is not correct:  it should be $$\binom{16}{7} + \binom{12}{7},$$ because the choice of an all-male group is exclusive of the choice of an all-female group.   Therefore, the number of such choices is the sum, not the product, of the number of choices within each sex. 
For a simpler example, suppose I have $3$ red balls labeled $R_1, R_2, R_3$ and $2$ blue balls labeled $B_1, B_2$.  How many ways can I choose $2$ balls of the same color?   There are $\binom{3}{2} = 3$ ways to choose two red balls.   There is only $\binom{2}{2} = 1$ way to choose two blue balls.   There should be $3 + 1 = 4$ ways to get two balls of the same color.