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-

Answer: Option D

n(S) = number of ways of choosing 4 persons out of 9
$$={}^9{C}_4$$   $$=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}$$   = 126
n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) $$=\left({}^4{C}_2\times {}^5{C}_2\right)$$   $$=\left(\frac{4\times 3}{2\times 1}\times \frac{5\times 4}{2\times 1}\right)$$     = 60
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{60}{126}=\frac{10}{21}$$