Let `A={x_(1),x_(2),x_(3)....,x_(7)},B={y_(1)y_(2)y_(3)}`. The total number of functions `f:AtoB` that are onto and ther are exactly three elements x in A such that `f(x)=y_(2)`, is equal to

Options

A. `14.^(7)C_(2)`
B. `14.^(7)C_(3)`
C. `7.^(7)C_(2)`
D. `7.^(7)C_(3)`

Answer

To solve the problem, we need to find the total number of onto functions \( f: A \to B \) where \( A = \{x_1, x_2, x_3, x_4, x_5, x_6, x_7\} \) and \( B = \{y_1, y_2, y_3\} \), with the condition that exactly 3 elements in \( A \) map to \( y_2 \).

### Step-by-Step Solution:

1. **Select 3 Elements from A to Map to y2**:
We need to choose 3 elements from the 7 elements in set \( A \) that will map to \( y_2 \). The number of ways to select 3 elements from 7 is given by the combination formula:
\[ \binom{7}{3} \]

2. **Assign Remaining Elements to y1 and y3**:
After selecting 3 elements to map to \( y_2 \), we have \( 7 - 3 = 4 \) elements left in set \( A \). These 4 elements can be mapped to either \( y_1 \) or \( y_3 \). Each of the 4 remaining elements has 2 choices (either \( y_1 \) or \( y_3 \)), so the total number of ways to assign these 4 elements is:
\[ 2^4 \]

3. **Subtract Cases Where Not All y1 and y3 are Used**:
We need to ensure that the function is onto, meaning both \( y_1 \) and \( y_3 \) must be used at least once. We will subtract the cases where all 4 remaining elements map to either \( y_1 \) or \( y_3 \):
- If all 4 elements map to \( y_1 \), there is 1 way.
- If all 4 elements map to \( y_3 \), there is also 1 way.
Thus, we need to subtract these 2 cases from our total:
\[ 2^4 - 2 \]

4. **Calculate Total Number of Onto Functions**:
Now, we can combine our results to find the total number of onto functions:
\[ \text{Total Onto Functions} = \binom{7}{3} \times (2^4 - 2) \]

5. **Final Calculation**:
- Calculate \( \binom{7}{3} \):
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
- Calculate \( 2^4 - 2 \):
\[ 2^4 = 16 \quad \Rightarrow \quad 16 - 2 = 14 \]
- Now multiply:
\[ \text{Total Onto Functions} = 35 \times 14 = 490 \]

### Conclusion:
The total number of onto functions \( f: A \to B \) such that exactly 3 elements in \( A \) map to \( y_2 \) is **490**.