Hello, I am having trouble understanding the answer to the above question. The answer is 6!/3! and below is my working. I am having trouble arriving at the above answer and here is my approach. The question is to find the permutations of KEEPER minus all the duplicates. Therefore, to count the permutations intuitively, I follow the method below: For the first position there are 6 possibilities i.e. K or E or E or P or E or R but we want to eliminate repeats thus all cases obtained with the first, second and third E are the same and this boils down to one case. Examining that one case closely where the word starts with E we can weed out all the repeats by dividing by 3! since there are 3 positions to be filled by E and all possible permutations are 3!. Thus for that one case where the starting letter is E we have number of arrangements = 5!/3!. Similarly for the words starting with K, P and R we get 5!/3! by the same reasoning as (K or P or R and (6-1)! possibilities for the remaining letters of that word). Thus the total is 5!/3! (for E) + 5!/3! (for K) + 5!/3! (for P) + 5!/3! (for R). This is equal to 4.5!/3! which is incorrect according to the answer. What am I missing. Sorry for such a long post. Thanks in advance for your replies!