let A={1,2,3,4,5}.If 'f' is bijection from A to Asuch that f(k) is not equal to k, where k belongs to A,then the possible number of such functions is

To determine the number of bijections \( f \) from the set \( A = \{1, 2, 3, 4, 5\} \) to itself such that \( f(k) \neq k \) for all \( k \in A \), we are essentially looking for the number of derangements of the set \( A \). A derangement is a permutation of a set where none of the elements appear in their original position.

Understanding Derangements

Derangements can be calculated using a specific formula. For a set of size \( n \), the number of derangements \( !n \) can be computed using the following recursive relation:

• Base cases: \( !0 = 1 \) and \( !1 = 0 \)
• Recursive formula: \( !n = (n - 1) \times (!n-1 + !n-2) \)

Calculating Derangements for Set A

In our case, we have \( n = 5 \). We can calculate \( !5 \) using the recursive formula:

• First, we need \( !2 \) and \( !3 \):
• For \( n = 2 \): \( !2 = 1 \) (the only derangement is \( (2, 1) \))
• For \( n = 3 \): \( !3 = 2 \) (the derangements are \( (2, 3, 1) \) and \( (3, 1, 2) \))
• Now, calculate \( !4 \):
• Using the formula: \( !4 = 3 \times (!3 + !2) = 3 \times (2 + 1) = 9 \)
• Finally, calculate \( !5 \):
• Using the formula: \( !5 = 4 \times (!4 + !3) = 4 \times (9 + 2) = 4 \times 11 = 44 \)

Final Result

Thus, the total number of bijections \( f \) from the set \( A \) to itself, where \( f(k) \neq k \) for all \( k \in A \), is \( 44 \). This means there are 44 different ways to rearrange the elements of set \( A \) such that no element remains in its original position.