There are four letters and four envelopes and exactly one letter is to be put in exactly one envelope with the correct address. If the letters are randomly inserted into the envelopes, then consider the following statements: 1. It is possible that exactly one letter goes into an incorrect envelope. 2. There are only six ways in which only two letters can go into the correct envelopes. Which of the statements given above is/are correct?

Statement 1: It is possible that exactly one letter goes into an incorrect envelope. Let's assume there are 4 letters (L1, L2, L3, L4) and 4 corresponding correct envelopes (E1, E2, E3, E4). If exactly one letter goes into an incorrect envelope, say L1 goes into E2 (which is incorrect for L1). This means L1 is not in E1. For the other three letters (L2, L3, L4) to be correct, they must be in their respective envelopes: L2 in E2, L3 in E3, L4 in E4. However, L1 is already in E2. This means L2 cannot be in E2. This creates a contradiction: if L1 is in E2, then L2 cannot be in E2 (and thus cannot be correct). Therefore, it is impossible for exactly one letter to be in an incorrect envelope. If one letter is incorrect, it displaces another letter, making at least two letters incorrect. So, Statement 1 is incorrect.

Statement 2: There are only six ways in which only two letters can go into the correct envelopes. If exactly two letters go into the correct envelopes, then the remaining two letters must go into incorrect envelopes. First, choose which two letters go into the correct envelopes. This can be done in C(4, 2) ways (4 choose 2). C(4, 2) = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6 ways. Let's say L1 and L2 are the letters that go into their correct envelopes (L1 in E1, L2 in E2). Now, the remaining two letters (L3 and L4) must go into the remaining two envelopes (E3 and E4) such that both are incorrect. This is a derangement of 2 items. For L3 and L4 to be incorrect in E3 and E4: L3 must go into E4 (not E3). L4 must go into E3 (not E4). There is only 1 way for this to happen (a derangement of 2 items is 1). So, the total number of ways is C(4, 2) * 1 = 6 * 1 = 6 ways. So, Statement 2 is correct.

Based on the analysis, only Statement 2 is correct.

The final answer is B