Seven different objects must be divided among three people. In how many ways can this be done if at least one of them gets exactly 1 object?

Options

A. 2484
B. 1218
C. 729
D. None of these

Answer

To solve the problem of distributing 7 different objects among 3 people such that at least one person gets exactly 1 object, we can break it down into two cases:

1. **Case 1**: Exactly one person gets 1 object, and the other two people share the remaining 6 objects.
2. **Case 2**: Exactly two people get 1 object each, and the third person gets the remaining 5 objects.

### Step-by-step Solution:

**Step 1: Case 1 - One person gets exactly 1 object**

- Choose 1 object out of 7 to give to the first person. This can be done in \( \binom{7}{1} \) ways.
- Now, we have 6 objects left to distribute among the remaining 2 people. The distribution can be done in several ways:
- One person gets all 6 objects, and the other gets none: \( \binom{6}{6} \)
- One person gets 5 objects, and the other gets 1 object: \( \binom{6}{5} \)
- One person gets 4 objects, and the other gets 2 objects: \( \binom{6}{4} \)
- One person gets 3 objects, and the other gets 3 objects: \( \binom{6}{3} \)

- The total ways to distribute the 6 objects can be calculated as follows:
\[ \text{Total for Case 1} = \binom{7}{1} \left( \binom{6}{6} + \binom{6}{5} \cdot 2! + \binom{6}{4} \cdot 3! + \binom{6}{3} \cdot 2! \cdot 2! \right) \]

Calculating each term:
- \( \binom{7}{1} = 7 \)
- \( \binom{6}{6} = 1 \)
- \( \binom{6}{5} = 6 \) (and \( 2! = 2 \))
- \( \binom{6}{4} = 15 \) (and \( 3! = 6 \))
- \( \binom{6}{3} = 20 \) (and \( 2! \cdot 2! = 4 \))

Putting it all together:
\[ \text{Total for Case 1} = 7 \left( 1 + 6 \cdot 2 + 15 \cdot 6 + 20 \cdot 4 \right) = 7 \left( 1 + 12 + 90 + 80 \right) = 7 \cdot 193 = 1351 \]

**Step 2: Case 2 - Two people get exactly 1 object each**

- Choose 2 objects out of 7 to give to two different people. This can be done in \( \binom{7}{2} \) ways.
- The remaining 5 objects will go to the third person.

Calculating:
\[ \text{Total for Case 2} = \binom{7}{2} = 21 \]

**Step 3: Combine the results from both cases**

Now, we add the totals from both cases to find the overall number of ways:
\[ \text{Total Ways} = \text{Total for Case 1} + \text{Total for Case 2} = 1351 + 21 = 1372 \]

### Final Answer:
The total number of ways to distribute the 7 different objects among 3 people such that at least one person gets exactly 1 object is **1372**.