If number of ways in which 7 different balls can be distributed into 4 boxes, so that no box remains empty is 48`lamda`, the value of `lamda ` is

Options

A. 231
B. 331
C. 175
D. 531

Answer

To solve the problem of distributing 7 different balls into 4 boxes such that no box remains empty, we can break it down into several cases based on how the balls can be distributed.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have 7 different balls and 4 boxes. We need to find the number of ways to distribute these balls into the boxes such that no box is empty.

2. **Identifying Cases**: We can distribute the balls in the following ways:
- **Case 1**: 1 ball in each of the first 3 boxes and 4 balls in the 4th box (1, 1, 1, 4).
- **Case 2**: 1 ball in each of the first 2 boxes, 2 balls in the 3rd box, and 3 balls in the 4th box (1, 1, 2, 3).
- **Case 3**: 1 ball in one box and 2 balls in each of the other 3 boxes (1, 2, 2, 2).

3. **Calculating Case 1**:
- Choose 1 ball for Box 1: \( \binom{7}{1} \)
- Choose 1 ball for Box 2: \( \binom{6}{1} \)
- Choose 1 ball for Box 3: \( \binom{5}{1} \)
- The remaining 4 balls go into Box 4.
- Since the first 3 boxes are identical, we divide by \(3!\) (for the arrangements of boxes).
- Additionally, since all 4 boxes can be interchanged, we multiply by \(4!\).

\[ C_1 = \binom{7}{1} \cdot \binom{6}{1} \cdot \binom{5}{1} \cdot \frac{4!}{3!} = 7 \cdot 6 \cdot 5 \cdot 4 = 840 \]

4. **Calculating Case 2**:
- Choose 1 ball for Box 1: \( \binom{7}{1} \)
- Choose 1 ball for Box 2: \( \binom{6}{1} \)
- Choose 2 balls for Box 3: \( \binom{5}{2} \)
- The remaining 3 balls go into Box 4.
- Since 2 boxes have identical counts, we divide by \(2!\) and multiply by \(4!\).

\[ C_2 = \binom{7}{1} \cdot \binom{6}{1} \cdot \binom{5}{2} \cdot \frac{4!}{2!} = 7 \cdot 6 \cdot 10 \cdot 12 = 5040 \]

5. **Calculating Case 3**:
- Choose 1 ball for Box 1: \( \binom{7}{1} \)
- Choose 2 balls for Box 2: \( \binom{6}{2} \)
- Choose 2 balls for Box 3: \( \binom{4}{2} \)
- The remaining 2 balls go into Box 4.
- Since all 3 boxes have identical counts, we divide by \(3!\) and multiply by \(4!\).

\[ C_3 = \binom{7}{1} \cdot \binom{6}{2} \cdot \binom{4}{2} \cdot \frac{4!}{3!} = 7 \cdot 15 \cdot 6 \cdot 4 = 2520 \]

6. **Total Ways**: Now we sum the total ways from all cases.
\[ \text{Total} = C_1 + C_2 + C_3 = 840 + 5040 + 2520 = 8400 \]

7. **Finding Lambda**: According to the problem, this total is equal to \(48\lambda\).
\[ 8400 = 48\lambda \implies \lambda = \frac{8400}{48} = 175 \]

### Final Answer:
The value of \( \lambda \) is \( 175 \).