In how any ways can 8 different books be distributed among 3 students if each receives at least 2 books?

Options

A. 490
B. 980
C. 2940
D. 5880

Answer

To solve the problem of distributing 8 different books among 3 students such that each student receives at least 2 books, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Determine the Distribution Cases**:
Since each student must receive at least 2 books, we can consider the following cases for the distribution of books:
- Case 1: Student 1 gets 2 books, Student 2 gets 2 books, and Student 3 gets 4 books.
- Case 2: Student 1 gets 2 books, Student 2 gets 3 books, and Student 3 gets 3 books.

2. **Calculate for Case 1**:
- For the first case, we need to choose 2 books for Student 1 from the 8 books. The number of ways to do this is given by \( \binom{8}{2} \).
- After choosing 2 books for Student 1, we have 6 books left. We now choose 2 books for Student 2 from these 6 books, which can be done in \( \binom{6}{2} \) ways.
- The remaining 4 books will automatically go to Student 3, which can be done in \( \binom{4}{4} \) ways (which is 1 way).
- Since there are 3 students, we multiply by the number of ways to assign these distributions to the students, which is 3! (since any of the three students can receive any of the distributions).

Thus, the total number of ways for Case 1 is:
\[ \text{Ways for Case 1} = \binom{8}{2} \times \binom{6}{2} \times \binom{4}{4} \times 3! \]

3. **Calculate for Case 2**:
- For the second case, we choose 2 books for Student 1 from the 8 books, which is \( \binom{8}{2} \).
- After choosing 2 books for Student 1, we have 6 books left. We now choose 3 books for Student 2 from these 6 books, which can be done in \( \binom{6}{3} \) ways.
- The remaining 3 books will go to Student 3, which can be done in \( \binom{3}{3} \) ways (which is 1 way).
- Again, we multiply by the number of ways to assign these distributions to the students, which is 3!.

Thus, the total number of ways for Case 2 is:
\[ \text{Ways for Case 2} = \binom{8}{2} \times \binom{6}{3} \times \binom{3}{3} \times 3! \]

4. **Calculate Each Case**:
- For Case 1:
\[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]
\[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \]
\[ \text{Ways for Case 1} = 28 \times 15 \times 1 \times 6 = 2520 \]

- For Case 2:
\[ \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
\[ \text{Ways for Case 2} = 28 \times 20 \times 1 \times 6 = 3360 \]

5. **Total Number of Ways**:
Now, we add the number of ways from both cases:
\[ \text{Total Ways} = 2520 + 3360 = 5880 \]

### Final Answer:
Thus, the total number of ways to distribute 8 different books among 3 students such that each receives at least 2 books is **5880**.

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