To determine the number of ways to select 3 books from a group of 11 books, we need to use the combination formula. The combination formula is given by [tex]\({ }_n C _r = \frac{n!}{r!(n-r)!}\)[/tex], where [tex]\(n\)[/tex] is the total number of items to choose from, and [tex]\(r\)[/tex] is the number of items to be chosen.
Here, [tex]\(n = 11\)[/tex] and [tex]\(r = 3\)[/tex].
1. First, we identify [tex]\(n\)[/tex] and [tex]\(r\)[/tex]:
[tex]\[
n = 11, \quad r = 3
\][/tex]
2. Next, we substitute [tex]\(n\)[/tex] and [tex]\(r\)[/tex] into the combination formula:
[tex]\[
{ }_{11}C_{3} = \frac{11!}{3!(11-3)!}
\][/tex]
3. We simplify the denominator by calculating [tex]\(11 - 3\)[/tex]:
[tex]\[
11 - 3 = 8
\][/tex]
4. The formula now becomes:
[tex]\[
{ }_{11}C_{3} = \frac{11!}{3! \cdot 8!}
\][/tex]
5. We observe that [tex]\(11!\)[/tex] can be expanded and simplified to cancel out the [tex]\(8!\)[/tex] in the numerator and denominator:
[tex]\[
11! = 11 \times 10 \times 9 \times 8!
\][/tex]
[tex]\[
\frac{11 \times 10 \times 9 \times 8!}{3! \times 8!}
\][/tex]
6. The [tex]\(8!\)[/tex] in the numerator and the denominator cancel each other out:
[tex]\[
\frac{11 \times 10 \times 9}{3!}
\][/tex]
7. We calculate [tex]\(3!\)[/tex]:
[tex]\[
3! = 3 \times 2 \times 1 = 6
\][/tex]
8. Now, we have:
[tex]\[
\frac{11 \times 10 \times 9}{6}
\][/tex]
9. Performing the multiplication on the numerator:
[tex]\[
11 \times 10 = 110
\][/tex]
[tex]\[
110 \times 9 = 990
\][/tex]
10. We then divide by the denominator:
[tex]\[
\frac{990}{6} = 165
\][/tex]
Therefore, the number of ways to select 3 books from a group of 11 books is:
[tex]\[
165
\][/tex]
