In the word PONTIAN, where are 7 letters, in all; of them, letter N repeats twice;
other letters are unique.


Consider three cases, sequentially.



Case 1.  There is no "N" in a four-letter code.

         Then there are 5 different letters to use, (P, O, T, I, A) 
         and we can use any of these 5 letters in each of 4 positions.

         In all, there are  = 625 such 4-letter codes.



Case 2.  There is only one letter "N" in a four-letter code.

         Then there are 6 different letters to use, (P, O, T, I, A, N).
         We place "N" in any of 4 positions, and we can use any of 5 remaining letters 
         in the remaining 3 positions.

         It gives us  = 500 such 4-letter codes.



Case 3.  There are two letters "N" in a four-letter code, but these two "N" can not appear together.

         Then for these two "N" we have  - 3 =  = 2*3-3 = 6-3 = 3 possible places/locations 
         in the row of 4 positions.  (Here minus 3 means excluding cases when two "N" appears together).

         In two remaining positions, we can place any of 5 remaining letters.
         it gives the factor  = 25.

         In all, there are 4*25 = 100 such 4-letter codes.


Now, to get the answer, we should add 625 + 500 + 100 = 1225  possible 4-letter codes.