Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children is:

The total number of ways of selecting 4 persons out of 3 men, 2 women and 4 children can be calculated as $$^9C_4$$ because there are in total of 3 + 2 + 4 = 9 persons and out of them 4 are selected.

The number of ways in which out of the 4 persons selected, exactly 2 of them are children can be calculated as $$^5C_2\times\ ^4C_2$$ because out of the 4 children, 2 must be selected and out of the other (3 men + 2 women), 5 persons, 2 must be selected.

The chance that exactly 2 of them are children can be calculated as,

$$\dfrac{^5C_2\times\ ^4C_2}{^9C_4}\ =\ \dfrac{\frac{5\times\ 4}{2}\times\ \frac{4\ \times\ 3}{2}}{\frac{9\ \times\ 8\ \times\ 7\ \times\ 6}{24}}\ =\ \dfrac{10\ \times\ 6}{126}\ =\ \dfrac{10}{21}$$

Hence, the correct answer is option D.