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

Options

A. 276
B. `""^(13)C_(10)-""^(5)C_(3)`
C. `""^(5)C_(3).""^(8)C_(7)+""^(5)C_(4).""^(8)C_(6)+""^(5)C_(5).""^(8)C_(5)`
D. `""^(13)C_(10)-""^(5)C_(2)`

Answer

To solve the problem of how many ways a student can answer 10 out of 13 questions while ensuring that he answers at least 3 of the first 5 questions, we can break it down into cases based on how many questions he answers from the first 5.

### Step 1: Identify 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 3 questions from the first 5.
2. Case 2: The student answers 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 questions from the first 5
- The student needs to answer 10 questions in total, so he must answer \(10 - 3 = 7\) questions from the remaining 8 questions.
- The number of ways to choose 3 questions from the first 5 is given by \( \binom{5}{3} \).
- The number of ways to choose 7 questions from the remaining 8 is given by \( \binom{8}{7} \).

Calculating these:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
\[ \binom{8}{7} = \frac{8!}{7!(8-7)!} = 8 \]
Thus, the total ways for Case 1:
\[ \text{Total for Case 1} = \binom{5}{3} \times \binom{8}{7} = 10 \times 8 = 80 \]

#### Case 2: Answering 4 questions from the first 5
- The student must answer \(10 - 4 = 6\) questions from the remaining 8 questions.
- The number of ways to choose 4 questions from the first 5 is \( \binom{5}{4} \).
- The number of ways to choose 6 questions from the remaining 8 is \( \binom{8}{6} \).

Calculating these:
\[ \binom{5}{4} = 5 \]
\[ \binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]
Thus, the total ways for Case 2:
\[ \text{Total for Case 2} = \binom{5}{4} \times \binom{8}{6} = 5 \times 28 = 140 \]

#### Case 3: Answering all 5 questions from the first 5
- The student must answer \(10 - 5 = 5\) questions from the remaining 8 questions.
- The number of ways to choose 5 questions from the first 5 is \( \binom{5}{5} \).
- The number of ways to choose 5 questions from the remaining 8 is \( \binom{8}{5} \).

Calculating these:
\[ \binom{5}{5} = 1 \]
\[ \binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
Thus, the total ways for Case 3:
\[ \text{Total for Case 3} = \binom{5}{5} \times \binom{8}{5} = 1 \times 56 = 56 \]

### Step 3: Sum All Cases
Now, we sum the total ways from all cases:
\[ \text{Total ways} = \text{Total for Case 1} + \text{Total for Case 2} + \text{Total for Case 3} \]
\[ = 80 + 140 + 56 = 276 \]

### Final Answer
The total number of ways in which the student can answer the questions while ensuring at least 3 of the first 5 questions are answered is **276**.