To determine the number of different exams the teacher can make by choosing problems from a set of 12, we need to look at the combinations. Specifically, we want to know how many ways we can choose 11 problems out of 12, because by choosing 11 problems, the 12th problem is automatically excluded, and vice versa.
The number of ways to choose [tex]\( r \)[/tex] items from [tex]\( n \)[/tex] items without regard to order is given by the binomial coefficient, denoted as [tex]\( \binom{n}{r} \)[/tex]. The formula for combinations is:
[tex]\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\][/tex]
For this problem, [tex]\( n = 12 \)[/tex] and [tex]\( r = 11 \)[/tex], so we substitute these values into the formula:
[tex]\[
\binom{12}{11} = \frac{12!}{11!(12-11)!}
\][/tex]
Simplifying further:
[tex]\[
\binom{12}{11} = \frac{12!}{11! \times 1!}
\][/tex]
Using the property that [tex]\( k! = k \times (k-1)! \)[/tex], we get:
[tex]\[
\frac{12!}{11! \times 1!} = \frac{12 \times 11!}{11! \times 1} = 12
\][/tex]
So, the number of different exams the teacher can make by picking 11 problems out of 12 is:
[tex]\[
12
\][/tex]
Therefore, the correct answer is:
C. [tex]\(\frac{12!}{11!}\)[/tex] which simplifies to 12.
