A teacher has a set of 12 problems to use on math exam The makes different versions the by putting 10 questions each How many exams can make?

Let $n$ be the number of problems the teacher has, so $n=12$.

The teacher wants to make different versions of the exam by selecting 10 questions each.

The number of ways to choose 10 questions from 12 questions is given by the combination formula:

$$ \binom{n}{k} = \binom{12}{10} = \frac{12!}{10!(12-10)!} = \frac{12!}{10!2!} = \frac{12 \times 11}{2 \times 1} = 6 \times 11 = 66 $$

Therefore, the teacher can make 66 different exams.

The number of ways to choose 10 questions from 12 is given by the combination formula:

$$ \binom{12}{10} = \frac{12!}{10!(12-10)!} = \frac{12!}{10!2!} = \frac{12 \times 11}{2 \times 1} = 6 \times 11 = 66 $$

Thus, the teacher can make 66 different exams.

Final Answer: The final answer is $\boxed{66}$