To determine how many different combinations of 4 books can be selected from a reading list of 8 books, we will use the concept of combinations. Combinations deal with selecting a group of items where the order does not matter.
The formula for combinations is given by:
[tex]\[
C(n, k) = \frac{n!}{k!(n-k)!}
\][/tex]
where [tex]\( n \)[/tex] is the total number of items, [tex]\( k \)[/tex] is the number of items to choose, and [tex]\( ! \)[/tex] denotes factorial, which is the product of all positive integers up to that number.
Here, [tex]\( n = 8 \)[/tex] and [tex]\( k = 4 \)[/tex].
Substitute these values into the formula:
[tex]\[
C(8, 4) = \frac{8!}{4!(8-4)!}
\][/tex]
[tex]\[
C(8, 4) = \frac{8!}{4! \cdot 4!}
\][/tex]
Now, calculate the factorials:
[tex]\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320
\][/tex]
[tex]\[
4! = 4 \times 3 \times 2 \times 1 = 24
\][/tex]
Substitute these factorials back into the formula:
[tex]\[
C(8, 4) = \frac{40320}{24 \times 24} = \frac{40320}{576} = 70
\][/tex]
Thus, the number of different combinations of 4 books from a list of 8 books is [tex]\( 70 \)[/tex].
Therefore, the correct answer is:
D. 70
