Let A={1,2,3,4,5} and B={-2,-1,0,1,2,3,4,5}.
Onto functions from A to A such that `f(i) ne i` for all `i`, is

Options

A. (a)44
B. (b)120
C. (c)56
D. (d)76

Answer

To solve the problem of finding the number of onto functions from set A to itself such that \( f(i) \neq i \) for all \( i \), we can follow these steps:

### Step 1: Understand the Problem
We need to find the number of onto functions (also known as surjective functions) from set \( A = \{1, 2, 3, 4, 5\} \) to itself, with the condition that no element maps to itself.

### Step 2: Calculate Total Onto Functions
The total number of onto functions from a set of \( n \) elements to itself is given by \( n! \). For our set \( A \), which has 5 elements, the total number of onto functions is:
\[ 5! = 120 \]

### Step 3: Use Inclusion-Exclusion Principle
To count the number of functions where \( f(i) \neq i \) for all \( i \), we can use the principle of inclusion-exclusion.

1. **Let \( S \) be the set of all onto functions.**
2. **Define \( A_i \) as the set of functions where \( f(i) = i \).**

We want to find the size of the complement of the union of all \( A_i \):
\[ |S| - |A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5| \]

### Step 4: Calculate \( |A_i| \)
If \( f(i) = i \), the remaining 4 elements can map to any of the 5 elements, which gives us:
\[ |A_i| = 4! = 24 \]

### Step 5: Calculate \( |A_i \cap A_j| \)
If \( f(i) = i \) and \( f(j) = j \), then we have 3 elements left, giving us:
\[ |A_i \cap A_j| = 3! = 6 \]

### Step 6: Generalize for \( k \) Fixed Points
Using the inclusion-exclusion principle, we calculate:
\[ |A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5| = \sum |A_i| - \sum |A_i \cap A_j| + \sum |A_i \cap A_j \cap A_k| - \ldots \]

The general term for \( k \) fixed points is:
\[ |A_{i_1} \cap A_{i_2} \cap \ldots \cap A_{i_k}| = (5-k)! \]

### Step 7: Calculate Each Term
Using the binomial coefficient \( \binom{5}{k} \):
- For \( k = 1 \): \( \binom{5}{1} \cdot 4! = 5 \cdot 24 = 120 \)
- For \( k = 2 \): \( \binom{5}{2} \cdot 3! = 10 \cdot 6 = 60 \)
- For \( k = 3 \): \( \binom{5}{3} \cdot 2! = 10 \cdot 2 = 20 \)
- For \( k = 4 \): \( \binom{5}{4} \cdot 1! = 5 \cdot 1 = 5 \)
- For \( k = 5 \): \( \binom{5}{5} \cdot 0! = 1 \cdot 1 = 1 \)

### Step 8: Apply Inclusion-Exclusion
Now, substituting these values into the inclusion-exclusion formula:
\[ |A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5| = 120 - 60 + 20 - 5 + 1 = 76 \]

### Step 9: Calculate Functions with No Fixed Points
Finally, the number of onto functions such that \( f(i) \neq i \) for all \( i \) is:
\[ |S| - |A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5| = 120 - 76 = 44 \]

### Conclusion
Thus, the number of onto functions from \( A \) to \( A \) such that \( f(i) \neq i \) for all \( i \) is **44**.