 Hello and welcome to the session. Let us understand the calling problem today. Let S is equal to A, B, C and T is equal to 1, 2, 3. Find F inverse of the calling function F from S to T if it exists. F is equal to A, 3, B, 2, C, 1. Now let us write the solution. Given to us S is equal to A, B, C and T is equal to 1, 2, 3. Let F be a function from S to T. Therefore F is a function from A, B, C to 1, 2, 3. And given to us F is equal to A, 3, B, 2 and C, 1. Now F is 1, 1 since F of X1 is equal to F of X2 which implies X1 is equal to X2. That is each element of B, C has a unique image 1, 2, 3. Also each element of T has a unique bijection. Now since F is a bijection therefore F is invertible from A is equal to 3, F of B is equal to 2 and F of C is equal to 1. Which implies F inverse of 3 is equal to A, F inverse of 2 is equal to B and F inverse of 1 is equal to C. Which implies F inverse is equal to 3, A, 2, B and 1, C. Therefore required F inverse is equal to 3, A, 2, B and 1, C. I hope you understood this problem. Bye and have a nice day.