Question 1044811
The sequence of the first 13 leaps (or 14 arrangements) are as follows:

ABC --> BAC ( B leaps to form the next arrangement)
ABC --> ACB --> CAB --> CBA --> BCA -->  BAC  (C leaps to form next arrangement) 
ABC --> ACB --> CAB --> CBA --> BCA -->  BAC  (C leaps to form next arrangement)
...........................

(cycle repeats itself after 6 arrangements after the first two arrangements)


For any k leaps, there are k+1 arrangements. 
For 1991 leaps there are 1992 arrangements.  Subtract the first two arrangements to start the count for the other arrangements, 
to get 1992 - 2 = 1990.
1990/6 = 331 + 4/6 (leaving a remainder of 4), meaning the last leap arrangement is CBA, which is not the initial position ABC.