To determine how many different exams the teacher can make by selecting 10 questions out of a set of 12, we need to use the concept of combinations. In combinations, the order in which the problems are selected does not matter.
The number of ways to choose [tex]\( k \)[/tex] questions out of [tex]\( n \)[/tex] questions is given by the combination formula:
[tex]\[ C(n, k) = \frac{n!}{k!(n - k)!} \][/tex]
Here:
- [tex]\( n = 12 \)[/tex] (the total number of problems)
- [tex]\( k = 10 \)[/tex] (the number of problems in each exam)
Plugging the values into the formula, we get:
[tex]\[ C(12, 10) = \frac{12!}{10!(12 - 10)!} = \frac{12!}{10! \cdot 2!} \][/tex]
Calculating [tex]\( 12! \)[/tex]:
[tex]\[ 12! = 12 \times 11 \times 10! \][/tex]
The [tex]\( 10! \)[/tex] terms cancel out:
[tex]\[ \frac{12 \times 11 \times 10!}{10! \cdot 2!} = \frac{12 \times 11}{2!} \][/tex]
Calculating [tex]\( 2! \)[/tex]:
[tex]\[ 2! = 2 \][/tex]
So we have:
[tex]\[ \frac{12 \times 11}{2} = \frac{132}{2} = 66 \][/tex]
Therefore, the number of different exams the teacher can make is:
[tex]\[ \boxed{66} \][/tex]
So, the correct answer is:
B. 66
