Let's solve the given question step-by-step:

Problem Breakdown:

The group consists of 4 girls and 7 boys. We need to determine the number of ways to select a team of 5 members under three conditions:

(i) No girls

If no girls are in the team, all 5 members must be boys.

• Total boys = $7$
• Choose all 5 members from 7 boys: $\binom{7}{5} = \frac{7 \times 6}{2 \times 1} = 21$

So, the number of ways is: $\boxed{21}$

(ii) At least one boy and one girl

We calculate the total number of ways to form a team of 5 members from 4 girls and 7 boys and subtract the cases where there are only boys or only girls.

Step 1: Total number of ways to form a team of 5 from all 11 people (4 girls + 7 boys): $\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$

Step 2: Cases with only boys (calculated in part (i)): $\binom{7}{5} = 21$

Step 3: Cases with only girls: $\binom{4}{5} = 0 \quad \text{(not possible as there are only 4 girls)}.$

Step 4: Subtract cases with only boys or only girls: $462 - 21 - 0 = 441$

So, the number of ways is: $\boxed{441}$

(iii) At least three girls

We calculate the cases where the team has 3 girls, 4 girls, or 5 girls.

Case 1: 3 girls and 2 boys:

• Choose 3 girls from 4: $\binom{4}{3} = \frac{4}{1} = 4$
• Choose 2 boys from 7: $\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$
• Total for this case: $4 \times 21 = 84$

Case 2: 4 girls and 1 boy:

• Choose 4 girls from 4: $\binom{4}{4} = 1$
• Choose 1 boy from 7: $\binom{7}{1} = 7$
• Total for this case: $1 \times 7 = 7$

Case 3: 5 girls (not possible since there are only 4 girls): $\binom{4}{5} = 0$

Total: Add the valid cases: $84 + 7 + 0 = 91$

So, the number of ways is: $\boxed{91}$

Final Answers:

(i) No girls: $\boxed{21}$
(ii) At least one boy and one girl: $\boxed{441}$
(iii) At least three girls: $\boxed{91}$

Do you have any questions, or would you like me to explain a specific part further? 😊

Related Questions:

1. How many ways can a team of 6 members be formed under the same conditions?
2. What if the group had 5 girls and 8 boys? Recalculate for all parts.
3. How would the solution change if the team size were 4 instead of 5?
4. In how many ways can a team of 5 boys only be selected if there were 10 boys instead of 7?
5. What if the condition required "exactly 2 girls"? How would this be solved?

Tip:

When solving combinatorics problems, always break the problem into smaller cases and use the formula for combinations $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.