Atest consists of 10 problems and students are told to answer any 4 of these questions. in how many different ways can they choose the 4 questions?

In this item, it is not given that there has to be specific arrangement or order as to how the questions are going to be answered. Hence, the concept that can be used in order to answer this item is the concept of combination which states that, 
            nCr = n! / (r!)(n - r)!

This is read as "combination of n taken r". In this item, n = 10 and r = 4. Substituting the known vaues to the equation,
          10C4 = (10!) / (4!)(10 - 4)!
          10C4 = 210

Therefore, the number of which is which the questions can be answered is 210. 

The answer is 10 bud

B? ? i'm most likely wrong but i think it would be b