There are 3 parts A,B and C in a question paper of Math's, which includes 6,7 and 8 questions respectivelty. From these parts 3,4 and 5 questions respectively are to be solved. In how many ways can a student select the questions from these parts?

Answer

To solve the problem of how many ways a student can select questions from parts A, B, and C of a math question paper, we will use the concept of combinations. Let's break it down step by step.

### Step 1: Understand the problem
We have three parts of a question paper:
- Part A has 6 questions, and the student needs to select 3 questions.
- Part B has 7 questions, and the student needs to select 4 questions.
- Part C has 8 questions, and the student needs to select 5 questions.

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

### Step 3: Calculate combinations for each part
1. **For Part A:**
- Total questions \( n = 6 \)
- Questions to select \( r = 3 \)
\[ \text{Ways to select from A} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

2. **For Part B:**
- Total questions \( n = 7 \)
- Questions to select \( r = 4 \)
\[ \text{Ways to select from B} = \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

3. **For Part C:**
- Total questions \( n = 8 \)
- Questions to select \( r = 5 \)
\[ \text{Ways to select from C} = \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5! \cdot 3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

### Step 4: Calculate the total number of ways
To find the total number of ways to select the questions from all parts, we multiply the number of ways from each part:
\[ \text{Total ways} = \text{Ways from A} \times \text{Ways from B} \times \text{Ways from C} = 20 \times 35 \times 56 \]

Calculating this:
\[ 20 \times 35 = 700 \]
\[ 700 \times 56 = 39200 \]

### Final Answer
The total number of ways a student can select the questions from parts A, B, and C is **39200**.

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