There are two solutions for this problem. It depends if order matters or not. As asked in the problem, we have to show both solutions.
The first solution is for the condition that order does not matter. This is a combination problem:
nCr = n!/r!(n-r)!
where n=10 because there are 10 total students, and r=2 because you choose 2 students at a time
nCr = 10!/2!(10-2)! = 45 ways
The second solution is for the condition that order matters. This is a permutation problem.
nPr = n!/(n-r)! = 10!/(10-2)! = 90 ways
