A student has to answer 10 out of 13 questions in an examination. The number of ways in which he can answer if he must answer atleast 3 of the first five questions is `276` b. `267` c. `80` d. `1200`

Options

A. 276
B. 600
C. 840
D. 640

Answer

To solve the problem, we need to find the number of ways a student can answer 10 out of 13 questions, ensuring that at least 3 of the first 5 questions are answered. We will break this down into cases based on how many of the first 5 questions the student answers.

### Step 1: Define the Cases
We will consider three cases based on how many questions the student answers from the first 5 questions:
1. Case 1: The student answers exactly 3 questions from the first 5.
2. Case 2: The student answers exactly 4 questions from the first 5.
3. Case 3: The student answers all 5 questions from the first 5.

### Step 2: Calculate Each Case

#### Case 1: Answering 3 out of the first 5 questions
- The number of ways to choose 3 questions from the first 5 is given by \( \binom{5}{3} \).
- After answering 3 questions from the first 5, the student needs to answer 7 more questions from the remaining 8 questions (13 total - 5 chosen = 8 remaining).
- The number of ways to choose 7 questions from these 8 is given by \( \binom{8}{7} \).

So, the total for Case 1 is:
\[ \text{Total for Case 1} = \binom{5}{3} \times \binom{8}{7} \]

#### Case 2: Answering 4 out of the first 5 questions
- The number of ways to choose 4 questions from the first 5 is \( \binom{5}{4} \).
- The student now needs to answer 6 more questions from the remaining 8 questions.
- The number of ways to choose 6 questions from these 8 is \( \binom{8}{6} \).

So, the total for Case 2 is:
\[ \text{Total for Case 2} = \binom{5}{4} \times \binom{8}{6} \]

#### Case 3: Answering all 5 questions from the first 5
- The number of ways to choose all 5 questions from the first 5 is \( \binom{5}{5} \).
- The student now needs to answer 5 more questions from the remaining 8 questions.
- The number of ways to choose 5 questions from these 8 is \( \binom{8}{5} \).

So, the total for Case 3 is:
\[ \text{Total for Case 3} = \binom{5}{5} \times \binom{8}{5} \]

### Step 3: Combine All Cases
Now, we sum the totals from all three cases:
\[ \text{Total Ways} = \text{Total for Case 1} + \text{Total for Case 2} + \text{Total for Case 3} \]

### Step 4: Calculate the Binomial Coefficients
Now we calculate the values:
1. \( \binom{5}{3} = 10 \)
2. \( \binom{8}{7} = 8 \)
3. \( \binom{5}{4} = 5 \)
4. \( \binom{8}{6} = 28 \)
5. \( \binom{5}{5} = 1 \)
6. \( \binom{8}{5} = 56 \)

### Step 5: Substitute and Calculate
Now substituting these values back into the total:
- For Case 1: \( 10 \times 8 = 80 \)
- For Case 2: \( 5 \times 28 = 140 \)
- For Case 3: \( 1 \times 56 = 56 \)

Adding these together:
\[ \text{Total Ways} = 80 + 140 + 56 = 276 \]

### Final Answer
The number of ways the student can answer the questions is **276**.