A teacher has four books—A, B, C, and D—to assign to four students in any order he prefers. Set S includes all possible arrangements of the four books. Set X includes all possible arrangements when book C is chosen first and D is chosen second. Which notation is correct? If book C is chosen first and D is chosen second, in how many ways can the teacher assign the books?

Answer with explanation:The Number of Books Possessed by the teacher = A, B,C,DTotal possible arrangement of the books,in which order is Important,we will use the concept of PermutationS={[tex]_{4}^{4}\textrm{P}=\frac{4!}{(4-4)!}=\frac{4!}{0!}=4\times 3\times 2\times 1=24[/tex]So, there are 24 elements in set S.Number of Elements in the set X               =taking CD as a first element and Either A as third and B as fourth or B as third and A as fourth.=1+1=2 waysOr Keeping CD as one book, then there are 2 books left,which can be arranged as=1×2×1=2 waysSo, if book C is chosen first and D is chosen second, number of ways the teacher can assign the books      = 2 ways