Sure! Let's solve the problem step-by-step using the concept of combinations.
To determine the number of different combinations in which we can choose 4 books out of a total of 12 books, we use the combinations formula:
[tex]\[
{}_n C _r = \frac{n!}{r!(n-r)!}
\][/tex]
Here, [tex]\( n \)[/tex] is the total number of items to choose from, and [tex]\( r \)[/tex] is the number of items to choose. In this problem, [tex]\( n = 12 \)[/tex] and [tex]\( r = 4 \)[/tex]. Substituting these values into the formula, we get:
[tex]\[
{}_{12} C _4 = \frac{12!}{4!(12-4)!} = \frac{12!}{4! \cdot 8!}
\][/tex]
Now let's explain what this means. The notation [tex]\( 12! \)[/tex] (read as "12 factorial") represents the product of all positive integers up to 12. Similarly, [tex]\( 4! \)[/tex] and [tex]\( 8! \)[/tex] represent the products of all positive integers up to 4 and 8 respectively.
To simplify [tex]\( \frac{12!}{4! \cdot 8!} \)[/tex]:
- Notice that [tex]\( 12! = 12 \times 11 \times 10 \times 9 \times 8! \)[/tex].
- Therefore, we can cancel out [tex]\( 8! \)[/tex] in the numerator and denominator:
[tex]\[
\frac{12 \times 11 \times 10 \times 9 \times 8!}{4! \times 8!} = \frac{12 \times 11 \times 10 \times 9}{4!}
\][/tex]
Next, we calculate [tex]\( 4! \)[/tex]:
[tex]\[
4! = 4 \times 3 \times 2 \times 1 = 24
\][/tex]
So our expression simplifies to:
[tex]\[
\frac{12 \times 11 \times 10 \times 9}{24}
\][/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]
And then divide by the denominator:
[tex]\[
\frac{11880}{24} = 495
\][/tex]
Thus, the number of different combinations of 4 books chosen from a list of 12 books is:
[tex]\[
\boxed{495}
\][/tex]
