Question

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 1

Answer with explanation:

The Number of Books Possessed by the teacher = A, B,C,D

Total possible arrangement of the books,in which order is Important,we will use the concept of Permutation

S={$_{4}^{4}\textrm{P}=\frac{4!}{(4-4)!}=\frac{4!}{0!}=4\times 3\times 2\times 1=24$

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 ways

Or

Keeping CD as one book, then there are 2 books left,which can be arranged as

=1×2×1

=2 ways

So, if book C is chosen first and D is chosen second, number of ways the teacher can assign the books

     = 2 ways

Answer 2

Answer:

x, s, and then 2

Step-by-step explanation:

i got it wrong and gave me the answer.