Q3(ii):

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (ii) atleast 3 girls ?

Step 1: Analyzing the Selection Constraints

The total population is $9$ boys and $4$ girls. We are required to form a committee of $7$ members. The constraint is that the committee must contain at least 3 girls.

Given the constraint of having at least $3$ girls in a $7$-member committee, we identify the valid cases based on the number of girls ($G$) and boys ($B$):

• Case 1: $3$ girls and $4$ boys.
• Case 2: $4$ girls and $3$ boys.

[By the Fundamental Counting Principle and the definition of Combinations, where selecting $r$ items from $n$ is given by $^nC_r = \frac{n!}{r!(n-r)!}$]

Step 2: Calculating Combinations for Each Case

Case 1: Selecting 3 girls and 4 boys

Number of ways to select $3$ girls from $4$ available girls: $^4C_3 = \frac{4!}{3!(4-3)!} = 4$.

Number of ways to select $4$ boys from $9$ available boys: $^9C_4 = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.

Total for Case 1: $4 \times 126 = 504$.

Case 2: Selecting 4 girls and 3 boys

Number of ways to select $4$ girls from $4$ available girls: $^4C_4 = \frac{4!}{4!(4-4)!} = 1$.

Number of ways to select $3$ boys from $9$ available boys: $^9C_3 = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.

Total for Case 2: $1 \times 84 = 84$.

Step 3: Summation of Mutually Exclusive Cases

Since these cases are mutually exclusive, we apply the Addition Principle of counting:

Total Ways = (Ways for Case 1) + (Ways for Case 2)

Total Ways = $504 + 84 = 588$.

Final Solution: The total number of ways to form the committee is 588.