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

In general the real functions which satisfies the condition that fof(x) = x are

(i) f(x) = a – x where a is any real no. 

 

But here,  f: A → A is the function which should satisfy the condition fof(x) = x where A = {1, 2, 3, 4, 5}.  

 

So, we can have the following such functions 

(i)  f(x) = x. : –          f =  {(1,1), (2,2), (3,3), (4,4), (5,5)}

(ii) f(x) = 6 – x  : –           f(x)  = {(1,5), (2,4), (3,3), (4,2), (5,1)}

 

 

Pl. ignore this solution – it is incomplete

First note that the function is injective: f(x) = f(y) implies f(f(x)) = f(f(y)) which yields x = y.

 

We say that a number x is fixed by f if f(x) = x. 

 

Now, if f(a) = b, with a and b unequal, then f(b) = a. That means we can pair up numbers in the set that are not fixed. Since the given set has 5 numbers this means, we can only have an odd number of fixed numbers i.e. 1, 3, or 5.

 

Case 1: 5 fixed numbers. There is obviously only one such function

 

Case 2: 3 fixed numbers: We can choose these three numbers in 5C3 = 10 ways. Say they are 1,2,3. Notice that we must have f(4) = 5 and f(5)=4. So we have one such function for each choice of fixed numbers.

 

So, this case yields 10 functions

 

Case 3: One fixed number. The fixed number has 5 choices.

 

Lets say it is 1. Now, if we know, for instance, f(2) = 3, then f(3) = 2. and hence f(4) = 5 and f(5) = 4. That means, the function is decided by choice of f(2), which gives three options for f.

 

So this case yields 5 X 3 = 15 cases.

 

Thus, we have a total of 26 choices of f, such that f(f(x))=x