Twelve education students, in groups of four, are taking part in a student-teacher program. Mark cannot be in the first group because he will be arriving late.How many ways can the instructor choose the first group of four education students?Please explain step by step how to get the answer.

This is a combination problem.

Given:
12 students
3 groups consisting of 4 students.
Mark can't be in the first group.

The combination formula that I used is: n! / r!(n-r)!
where: n = number of choices ; r = number of people to be chosen.

This is the formula I used because the order is not important and repetition is not allowed. 

Since Mark can't be considered in the first group, the value of n would be 11 instead of 12. value of r is 4.

numerator: n! = 11! = 39,916,800
denominator: r!(n-r)! = 4!(11-4)! = 4!*7! = 120,960

Combination = 39,916,800 / 120,960 = 330

There are 330 ways that the instructor can choose 4 students for the first group