The number of surjections from A={1,2,... n}, `n ge 2,` onto B = {a,b} is

Options

A. `""^(n)P_2`
B. `2^n -2`
C. `2^n-1`
D. none of these

Answer

To find the number of surjective functions (or onto functions) from the set \( A = \{1, 2, \ldots, n\} \) (where \( n \geq 2 \)) onto the set \( B = \{a, b\} \), we can use the formula for the number of surjective functions from a set of \( m \) elements to a set of \( r \) elements.

### Step-by-Step Solution:

1. **Identify the sets**:
- Set \( A \) has \( n \) elements (where \( n \geq 2 \)).
- Set \( B \) has 2 elements: \( a \) and \( b \).

2. **Understand the formula for surjective functions**:
- The number of surjective functions from a set of \( m \) elements to a set of \( r \) elements is given by:
\[ S(m, r) = r! \cdot \sum_{j=0}^{r} (-1)^j \binom{r}{j} (r-j)^m \]
- In our case, \( m = n \) and \( r = 2 \).

3. **Apply the formula**:
- Substitute \( r = 2 \) into the formula:
\[ S(n, 2) = 2! \cdot \sum_{j=0}^{2} (-1)^j \binom{2}{j} (2-j)^n \]
- Since \( 2! = 2 \), we have:
\[ S(n, 2) = 2 \cdot \sum_{j=0}^{2} (-1)^j \binom{2}{j} (2-j)^n \]

4. **Calculate the summation**:
- Calculate each term in the summation:
- For \( j = 0 \):
\[ (-1)^0 \binom{2}{0} (2-0)^n = 1 \cdot 1 \cdot 2^n = 2^n \]
- For \( j = 1 \):
\[ (-1)^1 \binom{2}{1} (2-1)^n = -1 \cdot 2 \cdot 1^n = -2 \]
- For \( j = 2 \):
\[ (-1)^2 \binom{2}{2} (2-2)^n = 1 \cdot 1 \cdot 0^n = 0 \]

5. **Combine the results**:
- Now combine the results of the summation:
\[ \sum_{j=0}^{2} (-1)^j \binom{2}{j} (2-j)^n = 2^n - 2 + 0 = 2^n - 2 \]

6. **Final calculation**:
- Substitute back into the formula for \( S(n, 2) \):
\[ S(n, 2) = 2 \cdot (2^n - 2) = 2^{n+1} - 4 \]

### Conclusion:
The number of surjective functions from \( A \) onto \( B \) is:
\[ \boxed{2^{n} - 2} \]