If we select 3 persons randomly from a group, consisting of 4 men, 3 women and 2 boys, then the probability that 2 men are selected is:

Correct option is A

Given:
Total number of people: 4 men + 3 women + 2 boys = 9
Solution:
Total ways to select 3 people from 9:
This is a combination problem, calculated as ⁹C₃ = $$\frac{9!}{ 3! × 6!}$$​= 84 ways.
Ways to select 2 men from 4: This is ⁴C₂ =$$\frac{ 4!} { (2! ×2!)}$$​ = 6 ways.
Ways to select 1 person from the remaining 5 (3 women and 2 boys): This is ⁵C₁ = 5 ways.
Total ways to select 2 men and 1 other: 6 ways (for men) × 5 ways (for the other) = 30 ways.
Probability of selecting 2 men: $$\frac{\text{(Number of ways to select 2 men and 1 other)}}{\text{ (Total ways to select 3 people)}}$$​ = $$\frac{30} { 84}$$​​

= $$\frac{5} { 14}$$​​