Correct option is (c) 45
To elaborate: There are six places to be filled in the multiset using the n distinct elements. At least one element has to occur exactly twice and that would leave 4 more places in the multiset means that at most four elements can occur exactly once. Thus there are two mutually exclusive cases as follows: 1) Exactly one element occurs exactly twice and select this element in n ways. Fill up the remaining four spots using 5 distinct elements from the remaining n−1 elements in ^n-1C4 ways. 2) Exactly four elements that occur at least once each. Hence, the total number of ways to form the multiset is
^nC2 + n * ^n-1C4 = ^6C2 + 6 * ^6-1C4 = 45.