To determine the number of ways to choose 3 books from a group of 11 books, we use the combination formula, which is given by:
[tex]\[ _nC_r = \frac{n!}{r!(n-r)!} \][/tex]
Where:
- [tex]\( n \)[/tex] is the total number of items to choose from (in this case, 11 books).
- [tex]\( r \)[/tex] is the number of items to be chosen (in this case, 3 books).
Substituting the given values into the formula, we have:
[tex]\[ _{11}C_3 = \frac{11!}{3!(11-3)!} \][/tex]
We need to calculate [tex]\( 11! \)[/tex] (11 factorial), which is the product of all positive integers up to 11:
[tex]\[ 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
Next, we need to calculate [tex]\( 3! \)[/tex] (3 factorial):
[tex]\[ 3! = 3 \times 2 \times 1 \][/tex]
And, we need to calculate [tex]\( (11-3)! = 8! \)[/tex] (8 factorial):
[tex]\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
Substituting these factorials into the combination formula provides:
[tex]\[ _{11}C_3 = \frac{11!}{3! \cdot 8!} \][/tex]
By canceling out the common factorial terms in the numerator and the denominator, the expression simplifies to:
[tex]\[ _{11}C_3 = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} \][/tex]
Performing the multiplication in the numerator:
[tex]\[ 11 \times 10 \times 9 = 990 \][/tex]
And the multiplication in the denominator:
[tex]\[ 3 \times 2 \times 1 = 6 \][/tex]
Finally, dividing the results gives:
[tex]\[ \frac{990}{6} = 165 \][/tex]
Therefore, the number of ways to select 3 books out of 11 is:
[tex]\[ \boxed{165} \][/tex]
