(i) In how many ways a committee of 4 members out of 5 gents and 6 ladies can be formed if there is at least one lady in the committee?
(ii) From a class of 12 boys and 8 girls, 10 students are to be chosen for a competition, including at least 4 boys and 4 girls. The two boys who won the prizes last year should be included. In how many was can this selection be made?

Answer

### Step-by-Step Solution

#### Part (i)

**Problem Statement:** In how many ways can a committee of 4 members be formed from 5 gents and 6 ladies if there is at least one lady in the committee?

1. **Identify the Total Members:**
- Gents = 5
- Ladies = 6
- Total members = 5 + 6 = 11

2. **Calculate Total Ways to Form a Committee of 4:**
- The total ways to form a committee of 4 members from 11 people (5 gents + 6 ladies) is given by the combination formula:
\[ \text{Total ways} = \binom{11}{4} \]
- Calculate:
\[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]

3. **Calculate Ways with No Ladies (All Gents):**
- The number of ways to select 4 gents from 5:
\[ \text{Ways with no ladies} = \binom{5}{4} = 5 \]

4. **Calculate Ways with At Least One Lady:**
- To find the number of ways with at least one lady, subtract the ways with no ladies from the total ways:
\[ \text{Ways with at least one lady} = \text{Total ways} - \text{Ways with no ladies} = 330 - 5 = 325 \]

5. **Final Answer for Part (i):**
- The number of ways to form a committee of 4 members with at least one lady is **325**.

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#### Part (ii)

**Problem Statement:** From a class of 12 boys and 8 girls, 10 students are to be chosen for a competition, including at least 4 boys and 4 girls. The two boys who won the prizes last year should be included. In how many ways can this selection be made?

1. **Identify Fixed Members:**
- 2 boys (who won last year) are already included. Thus, we need to select 8 more students.

2. **Determine Remaining Students:**
- Remaining boys = 12 - 2 = 10
- Girls = 8

3. **Possible Cases for Selection:**
- We need to consider cases based on the number of boys and girls selected, ensuring at least 4 boys and 4 girls:
- **Case 1:** 2 additional boys and 6 girls
- **Case 2:** 3 additional boys and 5 girls
- **Case 3:** 4 additional boys and 4 girls

4. **Calculate Each Case:**
- **Case 1:** 2 boys and 6 girls
\[ \text{Ways} = \binom{10}{2} \times \binom{8}{6} = 45 \times 28 = 1260 \]

- **Case 2:** 3 boys and 5 girls
\[ \text{Ways} = \binom{10}{3} \times \binom{8}{5} = 120 \times 56 = 6720 \]

- **Case 3:** 4 boys and 4 girls
\[ \text{Ways} = \binom{10}{4} \times \binom{8}{4} = 210 \times 70 = 14700 \]

5. **Total Ways:**
- Add the number of ways from all cases:
\[ \text{Total ways} = 1260 + 6720 + 14700 = 22680 \]

6. **Final Answer for Part (ii):**
- The total number of ways to select the students for the competition is **22680**.

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