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Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade.

(a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade?

(b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment.

(c) Compute the limiting probability as n approaches infinity in each case.

Think of this situation as having n spaces, n numbers and n alphabets.
Each space is initially associated with a number and an alphabet. The alphabets and numbers are then redistributed among the spaces. This can be compared to permutating the alphabets while separately permutating the numbers. There are n!n! possible ways to do this.

a)So your question is, in how many such permutations, none of the alphabets or letters retain their initial spaces. To fill the spaces with the alphabets, there are now (n-1) ways. Similarly, to arrange the numbers, there are (n-1) ways.
Hence, to simultaneously distribute one letter and one alphabet to each space such that no alphabet or letter is in its original space, there are
[(n-1)]^2 ways.

So, the probability that no student receives his own assignment or homework is

[(n-1)]^2/n!n!.

b)number of ways in which no student receives both their own quiz and their own homework assignment = number of ways in which the spaces can be filled such that no letter is in its original space + number of ways in which no number is in its original space - number of ways in which no number or letter is in its original space. This is 2.n!.(n-1) - (n-1)^2

Hence, the probability of the HWs and assignments being distributed in this way is

2.n!.(n-1) - (n-1)^2/n!n!.

(a) You have two permutations underneath each other, like this: $$begin{array}{rccccc}&1&2&3&4&5\hlinetext{Quiz}&2&4&1&3&5\text{H/w}&3&2&4&5&1end{array}$$

Both must be derangements, so the number of arrangements is $D_n^2$ out of $n!^2$.

(b) Adapting the inclusion-exclusion argument used for ordinary derangements, you get:

$$n!^2-{nchoose 1}(n-1)!^2+{nchoose 2}(n-2)!^2-...$$

successively counting arrangements in which at least $0,1,2,...$ students get back both of their own pieces of work. For $n=2,3,4,5,6$ the answers are $3, 26, 453, 11844, 439975$ (OEIS A089041).

(c) By factoring out $n!$ from the inclusion-exclusion formula for ordinary derangements, it is easy to show that the limit is $e^{-1}$; thus the limit in part (a) is $e^{-2}$. As for part (b), factoring out $n!^2$ from the formula above, the result is $1-tfrac1n+...$ which approaches $1$ as $ntoinfty$.