First we find the number of sequences of 4 letters with no two adjacent letters the same. There are 25 letters that can be used, all the letters except "I". Case 1. All 4 letters used are different. None of these have 2 adjacent letters (or any letters) the same. That's 25P4 = 303600. Case 2. Exactly 3 different letters are used. These have two letters the same. The only positions the two like letters can occur and not be adjacent are 1st and 3rd, 1st and 4th, and 2nd and 4th. Examples of sequences of these three forms are ABAC, ABCA, and ABCB. So every sequence must be of one the same 3 forms as those 3. There are 25 ways to put a letter where the A's are. There remain 24 ways to put a letter where the B's are. There remain 23 ways to put a letter where the C's are. That's (3)(25P3) = (3)(25)(24)(23) = 41400 Case 3. Exactly 2 different letters are used. So they can only be 2 pairs of like letters. To have no 2 like letters adjacent, one pair must come 1st and 3rd and the other pair 2nd and 4th. An example is ABAB. There are 25 ways to put a letter where the A's are. There remain 24 ways to put a different letter where the B's are. That's (25)(24) = 25P2 = 600 Total for all 3 cases = 303600+41400+600 = 345600 For each of those 345600 ways to choose the 4 letters, the 3 digits can be any integer from 100 to 999, inclusive. There are 999 integers from 1 to 999 inclusive, and we subtract 99 for the 99 integers from 1 to 99 inclusive, and get 999-99 = 900. So for each of the 345600 ways to choose the letters, there are 900 ways to choose the 3 digits. Answer (345600)(900) = 311040000 Edwin