`5` different objects are to be distributed among `3` persons such that no two persons get the same number of objects. Number of ways this can be done is,

Options

A. `60`
B. `90`
C. `120`
D. `150`

Explanation

To solve the problem of distributing 5 different objects among 3 persons such that no two persons receive the same number of objects, we can break it down into cases based on how many objects each person receives.

### Step-by-Step Solution:

1. **Identify the Distribution Cases**:
Since we have 3 persons and they cannot receive the same number of objects, the only possible distributions of 5 objects among 3 persons are:
- Case 1: One person gets 4 objects, another gets 1 object, and the last gets 0 objects (4, 1, 0).
- Case 2: One person gets 3 objects, another gets 2 objects, and the last gets 0 objects (3, 2, 0).

2. **Calculate for Case 1 (4, 1, 0)**:
- Choose 4 objects out of 5 to give to the first person. The number of ways to choose 4 objects from 5 is given by \( \binom{5}{4} \).
- The remaining object will go to the second person. There is only 1 way to choose 1 object from 1, which is \( \binom{1}{1} \).
- The third person gets 0 objects, which is a fixed case.
- We also need to consider the arrangement of the persons. Since there are 3 persons, they can be arranged in \( 3! \) ways.

Therefore, the total for Case 1 is:
\[ \text{Total for Case 1} = \binom{5}{4} \times \binom{1}{1} \times 3! = 5 \times 1 \times 6 = 30 \]

3. **Calculate for Case 2 (3, 2, 0)**:
- Choose 3 objects out of 5 to give to the first person. The number of ways to choose 3 objects from 5 is \( \binom{5}{3} \).
- Choose 2 objects from the remaining 2 for the second person, which is \( \binom{2}{2} \).
- The third person gets 0 objects.
- Again, the arrangement of the persons is \( 3! \).

Therefore, the total for Case 2 is:
\[ \text{Total for Case 2} = \binom{5}{3} \times \binom{2}{2} \times 3! = 10 \times 1 \times 6 = 60 \]

4. **Combine the Results**:
Now, we add the totals from both cases to get the final answer:
\[ \text{Total Ways} = \text{Total for Case 1} + \text{Total for Case 2} = 30 + 60 = 90 \]

Thus, the total number of ways to distribute 5 different objects among 3 persons such that no two persons receive the same number of objects is **90**.