To find the number of ways to choose 8 books from 12 books where the order does not matter, we can use the combination formula. The number of combinations of \( n \) items taken \( r \) at a time is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
In this scenario, \( n = 12 \) and \( r = 8 \). We need to calculate \( \binom{12}{8} \).
Using the combination formula:
\[ \binom{12}{8} = \frac{12!}{8!(12-8)!} = \frac{12!}{8! \cdot 4!} \]
We can simplify this expression. The calculation for \( 12! \) can be done as follows:
\[ 12! = 12 \times 11 \times 10 \times 9 \times 8! \]
Now substituting this back into our combination formula:
\[ \binom{12}{8} = \frac{12 \times 11 \times 10 \times 9 \times 8!}{8! \cdot 4!} \]
The \( 8! \) terms cancel out:
\[ \binom{12}{8} = \frac{12 \times 11 \times 10 \times 9}{4!} \]
Now, we compute \( 4! \):
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
So our expression becomes:
\[ \binom{12}{8} = \frac{12 \times 11 \times 10 \times 9}{24} \]
Next, we calculate the numerator:
\[ 12 \times 11 = 132 \] \[ 132 \times 10 = 1320 \] \[ 1320 \times 9 = 11880 \]
Now, divide by \( 24 \):
\[ \frac{11880}{24} = 495 \]
Thus, the number of ways to choose 8 books from 12 is
\[ \boxed{495} \]