A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has :
(i) no girl ?
(ii) at least one boy and one girl ?
(iii) at least 3 girls ?

Answer

To solve the problem, we will break it down into three parts as per the requirements of the question.

### Given:
- Number of girls = 4
- Number of boys = 7
- Total members to select = 5

### (i) Number of ways to select a team of 5 members with no girls:

Since we want a team with no girls, we can only select from the boys.

**Step 1:** Calculate the number of ways to choose 5 boys from 7 boys.

Using the combination formula:
\[ \text{Number of ways} = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
where \( n \) is the total number of boys and \( r \) is the number of boys to choose.

Here, \( n = 7 \) and \( r = 5 \):
\[ \text{Number of ways} = \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5! \cdot 2!} = \frac{7 \times 6}{2 \times 1} = 21 \]

**Answer for part (i):** 21 ways.

### (ii) Number of ways to select a team of 5 members with at least one boy and one girl:

To find the number of ways to select at least one boy and one girl, we can use the total number of ways to select any 5 members and subtract the cases where there are only boys or only girls.

**Step 1:** Calculate the total number of ways to select any 5 members from 11 (4 girls + 7 boys):
\[ \text{Total ways} = \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 462 \]

**Step 2:** Subtract the cases where there are only boys (calculated in part (i)):
- Only boys = 21 ways (from part (i))

**Step 3:** There are no cases with only girls since we cannot select 5 girls (only 4 available).

Thus, the number of ways with at least one boy and one girl:
\[ \text{Ways with at least one boy and one girl} = 462 - 21 = 441 \]

**Answer for part (ii):** 441 ways.

### (iii) Number of ways to select a team of 5 members with at least 3 girls:

We can have the following combinations of girls and boys:
1. 3 girls and 2 boys
2. 4 girls and 1 boy

**Case 1:** Selecting 3 girls and 2 boys:
- Number of ways to choose 3 girls from 4:
\[ \binom{4}{3} = 4 \]
- Number of ways to choose 2 boys from 7:
\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]
- Total ways for this case:
\[ \text{Total for Case 1} = 4 \times 21 = 84 \]

**Case 2:** Selecting 4 girls and 1 boy:
- Number of ways to choose 4 girls from 4:
\[ \binom{4}{4} = 1 \]
- Number of ways to choose 1 boy from 7:
\[ \binom{7}{1} = 7 \]
- Total ways for this case:
\[ \text{Total for Case 2} = 1 \times 7 = 7 \]

**Step 4:** Add the totals from both cases:
\[ \text{Total ways with at least 3 girls} = 84 + 7 = 91 \]

**Answer for part (iii):** 91 ways.

### Summary of Answers:
- (i) 21 ways
- (ii) 441 ways
- (iii) 91 ways