Question 205844
We have five subjects and we are dealing with combinations, not permutations here. We know that if we have five as a subset, there is 1 combination. 5C5. 4 as a subset, leaves us with 5 combinations... 5C4. 


Notice how we are working with combinations in the problem. We want sets of 5, 4, 3, 2, 1, and 0 (or empty set).


With combinations, we need only add 5C5 + 5C4 + 5C3 + 5C2 + 5C1 + 5C0.


You can solve a combination based off the definition, which states that nCr= n!/(n-r)!, where ! denotes factorial.


So, 5C5 + 5C4 + 5C3 + 5C2 + 5C1 + 5C0 = 1 + 5 + 10 + 10 + 5 + 1 = 32. Your answer is "B".