There are 8 questions in a paper and a student have to attempt 5 questions. How many ways it can be done if
(i) there is no restriction?
(ii) first two questions are compulsory.
(iii) at least 3 questions are compulsory out of first 5 questions?

Answer

To solve the given problem step by step, we will break it down into three parts as specified in the question.

### Part (i): No restriction
We need to find out how many ways a student can choose 5 questions out of 8 without any restrictions.

1. **Identify the total number of questions and the number of questions to choose**:
- Total questions (n) = 8
- Questions to attempt (r) = 5

2. **Use the combination formula**:
The number of ways to choose r items from n items is given by the formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

3. **Substituting the values**:
\[ \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5! \cdot 3!} \]

4. **Calculating factorials**:
\[ 8! = 8 \times 7 \times 6 \times 5! \quad \text{(we can cancel } 5! \text{)} \]
\[ \binom{8}{5} = \frac{8 \times 7 \times 6}{3!} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56 \]

5. **Final answer for part (i)**:
The number of ways to choose 5 questions from 8 without any restrictions is **56**.

### Part (ii): First two questions are compulsory
Now, we need to find out how many ways a student can choose 5 questions if the first 2 questions are compulsory.

1. **Select the compulsory questions**:
- The first 2 questions are already chosen.

2. **Determine the remaining questions to choose**:
- We need to choose 3 more questions from the remaining 6 questions (questions 3 to 8).

3. **Use the combination formula**:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!} \]

4. **Calculating factorials**:
\[ \binom{6}{3} = \frac{6 \times 5 \times 4}{3!} = \frac{6 \times 5 \times 4}{6} = 5 \times 4 = 20 \]

5. **Final answer for part (ii)**:
The number of ways to choose 5 questions when the first 2 are compulsory is **20**.

### Part (iii): At least 3 questions are compulsory out of the first 5 questions
In this part, we need to find out how many ways a student can choose 5 questions such that at least 3 questions are chosen from the first 5 questions.

1. **Identify possible cases**:
- Case 1: Choose 3 questions from the first 5 and 2 from the remaining 3.
- Case 2: Choose 4 questions from the first 5 and 1 from the remaining 3.
- Case 3: Choose all 5 questions from the first 5.

2. **Calculating each case**:
- **Case 1**:
\[ \binom{5}{3} \times \binom{3}{2} = \frac{5!}{3! \cdot 2!} \times \frac{3!}{2! \cdot 1!} = 10 \times 3 = 30 \]
- **Case 2**:
\[ \binom{5}{4} \times \binom{3}{1} = \frac{5!}{4! \cdot 1!} \times \frac{3!}{1! \cdot 2!} = 5 \times 3 = 15 \]
- **Case 3**:
\[ \binom{5}{5} \times \binom{3}{0} = 1 \times 1 = 1 \]

3. **Totaling the cases**:
\[ \text{Total} = 30 + 15 + 1 = 46 \]

4. **Final answer for part (iii)**:
The number of ways to choose 5 questions such that at least 3 are from the first 5 questions is **46**.

### Summary of Answers:
- (i) 56 ways
- (ii) 20 ways
- (iii) 46 ways