3. In an examination a candidate is given the four answers to four questions but is not told which answer applies to which question. He is asked to write down each of the four answers next to its appropriate question.

b). show that there are 6 ways of getting just two answers in the correct places.

c). If a candidate guesses at random where the four answers are to go and X is the number of correct guesses he makes, draw up the probability distribution for X in tabular form.

this question has been asked before in an old post, with no answer. Balls in bags, 4 answers to 4 exam questions.

For b), is it $ ^4C_2$? it is just a matter of choosing which 2 of the 4 are correct. for the remaining 2, there is only one way that they can be wrong, that is when the 2 are swapped. so,$ ^4C_2$. but i myself am not very persuaded by this approach. wondering if there is a better way which can be applied to other similar cases (because i don't think i can use this approach to do other similar cases, like finding the number of ways of getting 5 answers correct if there are 11 questions and 11 answers. btw it would be cool if somebody can tell me how to do this too)

For c) i spend so much time trying this, and i just cannot find the number of ways for 0 correct answer and 1 correct answer! this is not just the simple arranging objects that i have been doing, this involves arranging objects into fixed spaces. fyi the answer for 0 correct spaces is 9 ways and 1 correct answer is 8 ways. i googled and found this reply(for 0 correct answers):

"This is just the number of permutations of the set {1,2,3,4} that change eachelement. Each permutation that moves all elements can be represented in the form (1 x y z) or (1 x)(y z), where x = (1), z = (y), etc.

So the # of possible permutations that change all elements is just the number of possible configurations of (x y z) = plus the number of configurations of (1 x)(y z) which is 3, for a total of 9 ways."

but i do not understand the notation he is using. somebody please explain this