To solve the problem, we need to determine how many different ways there are to choose 4 books from a reading list of 12 books. This is a combinations problem because the order in which the books are chosen does not matter.
We need to calculate the number of combinations of 12 items taken 4 at a time, denoted as [tex]\( C(12, 4) \)[/tex]. The formula for combinations is:
[tex]\[ C(n, k) = \frac{n!}{k!(n-k)!} \][/tex]
In our case, [tex]\( n = 12 \)[/tex] and [tex]\( k = 4 \)[/tex]. So we plug in these values into the formula:
[tex]\[ C(12, 4) = \frac{12!}{4!(12-4)!} \][/tex]
Simplify the formula step-by-step:
[tex]\[ C(12, 4) = \frac{12!}{4! \cdot 8!} \][/tex]
Since [tex]\( 12! \)[/tex] (12 factorial) is the product of all positive integers up to 12, it can be expanded as:
[tex]\[ 12! = 12 \times 11 \times 10 \times 9 \times 8! \][/tex]
Notice that [tex]\( 8! \)[/tex] appears in both the numerator and the denominator, so they cancel out:
[tex]\[ C(12, 4) = \frac{12 \times 11 \times 10 \times 9 \times 8!}{4! \times 8!} \][/tex]
Now we are left with:
[tex]\[ C(12, 4) = \frac{12 \times 11 \times 10 \times 9}{4!} \][/tex]
Next, calculate [tex]\( 4! \)[/tex] (4 factorial):
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \][/tex]
So the calculation becomes:
[tex]\[ C(12, 4) = \frac{12 \times 11 \times 10 \times 9}{24} \][/tex]
Now, perform the multiplication in the numerator:
[tex]\[ 12 \times 11 = 132 \][/tex]
[tex]\[ 132 \times 10 = 1320 \][/tex]
[tex]\[ 1320 \times 9 = 11880 \][/tex]
So, we have:
[tex]\[ C(12, 4) = \frac{11880}{24} \][/tex]
Finally, divide 11880 by 24:
[tex]\[ 11880 \div 24 = 495 \][/tex]
Thus, the number of different ways to choose 4 books from 12 is:
[tex]\[ \boxed{495} \][/tex]
Hence, the correct answer is:
[tex]\[ OB. 495 ways \][/tex]
