In an examination, a question pater consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a students select the questions ?

Answer

To solve the problem of how many ways a student can select questions from the examination paper, we will break it down step by step.

### Step 1: Understand the Problem
We have two parts of questions:
- Part I: 5 questions
- Part II: 7 questions

A student must attempt a total of 8 questions, with at least 3 questions from each part.

### Step 2: Determine the Possible Combinations
Given the requirement of selecting at least 3 questions from each part, we can have the following combinations of selections:
1. 3 questions from Part I and 5 questions from Part II
2. 4 questions from Part I and 4 questions from Part II
3. 5 questions from Part I and 3 questions from Part II

### Step 3: Calculate the Number of Ways for Each Combination

#### Combination 1: 3 from Part I and 5 from Part II
- The number of ways to choose 3 questions from Part I (5 questions):
\[ \text{Ways from Part I} = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

- The number of ways to choose 5 questions from Part II (7 questions):
\[ \text{Ways from Part II} = \binom{7}{5} = \binom{7}{2} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

- Total ways for this combination:
\[ \text{Total for Combination 1} = 10 \times 21 = 210 \]

#### Combination 2: 4 from Part I and 4 from Part II
- The number of ways to choose 4 questions from Part I:
\[ \text{Ways from Part I} = \binom{5}{4} = 5 \]

- The number of ways to choose 4 questions from Part II:
\[ \text{Ways from Part II} = \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

- Total ways for this combination:
\[ \text{Total for Combination 2} = 5 \times 35 = 175 \]

#### Combination 3: 5 from Part I and 3 from Part II
- The number of ways to choose 5 questions from Part I:
\[ \text{Ways from Part I} = \binom{5}{5} = 1 \]

- The number of ways to choose 3 questions from Part II:
\[ \text{Ways from Part II} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

- Total ways for this combination:
\[ \text{Total for Combination 3} = 1 \times 35 = 35 \]

### Step 4: Calculate the Total Number of Ways
Now, we add the total ways from all combinations:
\[ \text{Total Ways} = 210 + 175 + 35 = 420 \]

### Final Answer
The total number of ways a student can select the questions is **420**.

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