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}$

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.