A question has two parts. Part A and Part B, each containing 8 questions. If the students has to choose 6 from part A and 5 questions from part B, in how many ways can he choose the questions?

Options

A. 1268
B. 1788
C. 1024
D. 1568

Answer

To solve the problem of choosing questions from Part A and Part B, we can break it down into steps:

### Step 1: Understand the problem
We need to choose 6 questions from Part A, which has 8 questions, and 5 questions from Part B, which also has 8 questions.

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

### Step 3: Calculate combinations for Part A
For Part A, we need to choose 6 questions from 8:
\[ \binom{8}{6} = \frac{8!}{6!(8-6)!} = \frac{8!}{6! \cdot 2!} \]
Calculating this:
\[ = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]

### Step 4: Calculate combinations for Part B
For Part B, we need to choose 5 questions from 8:
\[ \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5! \cdot 3!} \]
Calculating this:
\[ = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

### Step 5: Use the principle of multiplication
Since the choices from Part A and Part B are independent, we multiply the number of ways to choose from each part:
\[ \text{Total ways} = \binom{8}{6} \times \binom{8}{5} = 28 \times 56 \]

### Step 6: Calculate the total
Now, we calculate:
\[ 28 \times 56 = 1568 \]

### Final Answer
The total number of ways a student can choose the questions is **1568**.

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