Let f : A-> A where A = { 1, 2, 3, 4, 5} such that ff(x) = x . How many such functions are possible

To determine how many functions \( f: A \to A \) exist such that \( ff(x) = x \) for all \( x \in A \), we first need to understand what the condition \( ff(x) = x \) implies. This condition means that applying the function \( f \) twice returns the original input. In other words, \( f \) is an **involution**, which is a function that is its own inverse.

Understanding Involutions

An involution has a few key properties:

• Each element in the set can either map to itself or to another element, such that the second application of the function returns the original element.
• Elements that map to each other must form pairs, while elements that map to themselves are fixed points.

Analyzing the Set A

Given the set \( A = \{ 1, 2, 3, 4, 5 \} \), we can categorize the elements based on how they can be paired or left as fixed points. The total number of elements is 5, which is odd. This means we can have:

• 0 pairs and 5 fixed points
• 1 pair and 3 fixed points
• 2 pairs and 1 fixed point

Counting the Functions

Now, let’s calculate the number of valid functions for each case:

Case 1: 0 pairs (5 fixed points)

In this scenario, every element maps to itself. There is exactly 1 function:

• f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4, f(5) = 5

Case 2: 1 pair (3 fixed points)

Here, we choose 2 elements to form a pair, and the remaining 3 will be fixed points. The number of ways to choose 2 elements from 5 is given by \( \binom{5}{2} \), which equals 10. Each choice of 2 elements can be paired in exactly 1 way. Thus, we have:

• 10 functions for this case.

Case 3: 2 pairs (1 fixed point)

In this case, we need to choose 4 elements to form 2 pairs, leaving 1 fixed point. The number of ways to choose 4 elements from 5 is \( \binom{5}{4} = 5 \). Once we have chosen 4 elements, we can pair them. The number of ways to pair 4 elements is given by \( \frac{4!}{2! \cdot 2!} = 3 \). Therefore, the total for this case is:

• 5 choices of fixed points × 3 ways to pair = 15 functions.

Summing It All Up

Now, we can add the number of functions from all cases:

• Case 1: 1 function
• Case 2: 10 functions
• Case 3: 15 functions

Thus, the total number of functions \( f: A \to A \) such that \( ff(x) = x \) is:

Total = 1 + 10 + 15 = 26

In conclusion, there are 26 different functions that satisfy the condition \( ff(x) = x \) for the set \( A = \{ 1, 2, 3, 4, 5 \} \).