Math trouble..someone help

I have to figure out this problem, Im really bad with math:/
How many ways can a teacher choose 2 students from a group of 4 students?
A.) If just 2 on a committee
B.) If for president and vice president of class

Someone please helpp(:

This type of problem involves combinations (or permutations, depending on if the order matters)

Imagine 2 chairs labeled chair 1 and chair 2. The students are A, B, C, and D. How can they be arranged?

AB
--C
--D
BA
--C
--D
CA
--B
--D
DA
--B
--C

That was a total of 12 ways. There were 4 "first" letters, and 3 "second" letters for each first. That meant there were 4 * 3 ways. What is the relationship between that and 4? In math, there's something called a factorial (written n!) which says multiply this number by every single whole number less than it until 1. With this kind of problem, the number of ways they can be arranged in 2 spots is 4!/(4-2)!. That's the same as 4*3.
If we had 3 chairs instead, it would be 4!/(4-3)!, which is (4*3*2*1)/1, and that's 24.
I don't understand what you mean by "just 2 on a committee", but for president and vice president that means that order matters, so the answer would be 12.

a= (1,2,) (1,3,) (1,4,) (2,3,) (2,4,) (3,4)
so 6 ways

b=1,2. 1,3. 1,4.
so thats 3 ways*4 students
so 12 ways