Let's break down question 2(b):

The problem involves selecting a committee of four people from a group of 5 boys and 6 girls, with the condition that "there must be at least two girls."

Step-by-step solution:

1. The committee has 4 members, and we are required to have at least 2 girls.
2. We will split this into three cases based on the number of girls in the committee:
   • Case 1: 2 girls and 2 boys.
   • Case 2: 3 girls and 1 boy.
   • Case 3: 4 girls and 0 boys.

Case 1: 2 girls and 2 boys

We select 2 girls from 6 and 2 boys from 5.

• Number of ways to select 2 girls from 6 = $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$
• Number of ways to select 2 boys from 5 = $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$

So, total ways for this case = $15 \times 10 = 150$

Case 2: 3 girls and 1 boy

We select 3 girls from 6 and 1 boy from 5.

• Number of ways to select 3 girls from 6 = $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
• Number of ways to select 1 boy from 5 = $\binom{5}{1} = 5$

So, total ways for this case = $20 \times 5 = 100$

Case 3: 4 girls and 0 boys

We select all 4 members as girls, so:

• Number of ways to select 4 girls from 6 = $\binom{6}{4} = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = 15$

So, total ways for this case = 15.

Final Total:

Add all the cases together:

• Total ways = $150 + 100 + 15 = 265$

Thus, the total number of ways to select the committee with at least 2 girls is 265.

Would you like further details on this problem or any additional assistance?

Here are 5 related questions to expand your understanding:

1. How do combinations differ from permutations, and why are combinations used in this problem?
2. What is the formula for combinations, and how does it work for large values?
3. How would the solution change if there had to be at least 3 girls on the committee?
4. Can you calculate how many ways to form a committee with only boys?
5. How many different committees could be formed if there were no restrictions?

Tip: When tackling counting problems, breaking them into cases often simplifies complex conditions.