Q.A committee of 3 is to be selected at random from a group of 4 men and 3 women. Find the probability that the committee consists of exactly 2 men and 1 woman.

Use combinations to count the favourable committees (2 men from 4, 1 woman from 3) against the total number of possible committees (any 3 from 7).

Step 1 — Find the total number of possible committees. Any 3 people are chosen from the total group of $4+3=7$: $\binom{7}{3} = \dfrac{7!}{3!\,4!} = 35$.

Step 2 — Find the number of favourable committees (exactly 2 men and 1 woman). Choose 2 men from 4: $\binom{4}{2} = 6$. Choose 1 woman from 3: $\binom{3}{1}=3$. By the multiplication principle of counting, the number of favourable committees is $6\times3=18$.

Step 3 — Apply the classical probability formula.

$P(\text{2 men, 1 woman}) = \dfrac{18}{35}$ …

Using permutations (order-sensitive counting) instead of combinations for both the numerator and denominator is a common error — since a committee has no internal order (being 'selected as a member' is the same regardless of the sequence of selection), combinations are the correct tool throughout, and as long as both numerator and denominator consistently use the same counti …