A man has 4 sons. There are 6 proposals of marriage for his sons. In how many ways can they select a proposal for their marriages, such that none of them marry with more than one girl?

Options

A. 180
B. 270
C. 360
D. none of these

Answer

To solve the problem of how many ways the man can select marriage proposals for his 4 sons from 6 proposals, ensuring that none of the sons marry more than one girl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the number of sons and proposals**:
- There are 4 sons (let's denote them as A, B, C, and D).
- There are 6 marriage proposals (denote them as P1, P2, P3, P4, P5, and P6).

2. **Choose a proposal for the first son**:
- The first son (A) can choose any of the 6 proposals.
- So, there are 6 choices for A.

3. **Choose a proposal for the second son**:
- After A has chosen a proposal, that proposal is no longer available for the other sons.
- Therefore, the second son (B) has 5 remaining proposals to choose from.
- So, there are 5 choices for B.

4. **Choose a proposal for the third son**:
- After A and B have chosen their proposals, 2 proposals are now taken.
- The third son (C) will have 4 proposals left to choose from.
- So, there are 4 choices for C.

5. **Choose a proposal for the fourth son**:
- After A, B, and C have chosen their proposals, 3 proposals are now taken.
- The fourth son (D) will have 3 proposals left to choose from.
- So, there are 3 choices for D.

6. **Calculate the total number of ways**:
- To find the total number of ways the sons can select their proposals, we multiply the number of choices for each son:
\[ \text{Total ways} = 6 \times 5 \times 4 \times 3 \]

7. **Perform the multiplication**:
- First, calculate \(6 \times 5 = 30\)
- Next, calculate \(30 \times 4 = 120\)
- Finally, calculate \(120 \times 3 = 360\)

8. **Conclusion**:
- Therefore, the total number of ways the 4 sons can select their marriage proposals is **360**.

### Final Answer:
The answer is **360**.

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