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

A. 12
B. 66
C. 132
D. [tex]$\frac{12!}{11!}$[/tex]



Answer :

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