If there are 30 onto, mapping from a set containing n elements to the set {0,1} then n equals

Options

A. 3
B. 5
C. 7
D. 2

Answer

To solve the problem of finding the value of \( n \) such that there are 30 onto mappings from a set containing \( n \) elements to the set \{0, 1\}, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the number of onto functions (or mappings) from a set \( A \) with \( n \) elements to a set \( B \) with 2 elements (specifically, \{0, 1\}).

2. **Use the Formula for Onto Functions**: The number of onto functions from a set of \( n \) elements to a set of \( m \) elements is given by:
\[ m^n - \binom{m}{1}(m-1)^n + \binom{m}{2}(m-2)^n - \ldots \]
For our case, \( m = 2 \).

3. **Plug in the Values**: Substitute \( m = 2 \) into the formula:
\[ \text{Number of onto functions} = 2^n - \binom{2}{1}(1^n) + \binom{2}{2}(0^n) \]
This simplifies to:
\[ 2^n - 2 \cdot 1^n + 0 = 2^n - 2 \]

4. **Set the Equation**: We know from the problem statement that there are 30 onto mappings:
\[ 2^n - 2 = 30 \]

5. **Solve for \( n \)**: Rearranging gives:
\[ 2^n = 30 + 2 = 32 \]
Now, we can express 32 as a power of 2:
\[ 2^n = 2^5 \]

6. **Equate the Exponents**: Since the bases are the same, we can equate the exponents:
\[ n = 5 \]

### Final Answer:
Thus, the value of \( n \) is \( 5 \).