Question

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?

Answer 1

The teacher has to take 10 questions at a time from 12 questions.

Total ways = 12C10 = 12!/(10!)(12-10)! = 12 x 11 / 2 = 66

The teacher can make 66 different exams

Answer 2

Answer:

66 ways

Step-by-step explanation:

Given that 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.

Available questions =12

Problems to be made out of these 12 = 10

Since order does not matter in the question paper having problems, this is a question of combination.

This can be done in 12C10 ways

$=\frac{12!}{2!10!}=66 ways$