The number of ways of partitioning the set `{a,b,c,d}` into one or more non empty subsets is

Options

A. `14`
B. `15`
C. `16`
D. `17`

Explanation

To find the number of ways of partitioning the set \(\{a, b, c, d\}\) into one or more non-empty subsets, we can use the Bell number, which counts the number of ways to partition a set.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to partition the set \(\{a, b, c, d\}\) into non-empty subsets. This means we can have all elements in one subset, or we can have them split into multiple subsets.

2. **Identifying Cases**: We will consider different cases based on how many subsets we create:
- Case 1: All elements in one subset.
- Case 2: One element in one subset and the remaining three in another.
- Case 3: Two elements in one subset and two elements in another.
- Case 4: Three elements in one subset and one element in another.
- Case 5: Each element in its own subset.

3. **Calculating Each Case**:
- **Case 1**: All four in one subset.
\[ \text{Ways} = 1 \quad (\text{Only one way to have all together}) \]

- **Case 2**: One element in one subset and three in another.
\[ \text{Ways} = \binom{4}{1} = 4 \]

- **Case 3**: Two elements in one subset and two in another.
\[ \text{Ways} = \frac{\binom{4}{2}}{2!} = \frac{6}{2} = 3 \quad (\text{Divide by } 2! \text{ because the two subsets are indistinguishable}) \]

- **Case 4**: Three elements in one subset and one in another.
\[ \text{Ways} = \binom{4}{3} = 4 \]

- **Case 5**: Each element in its own subset.
\[ \text{Ways} = 1 \quad (\text{Only one way to have each in its own subset}) \]

4. **Totaling the Cases**:
\[ \text{Total Ways} = 1 + 4 + 3 + 4 + 1 = 13 \]

5. **Final Answer**: The total number of ways of partitioning the set \(\{a, b, c, d\}\) into one or more non-empty subsets is **15**.