UPSEE Solved Model Paper 1

String is of 11 alphabets and no. of A’s are odd ⇒ Middle alphabet at 6th position must be A otherwise alphabets cannot be equally balanced.
After fixing A, there must be 2A’s and 3B’s in positions {1-5} and {7-11}. Any arrangement on one side fixes the arrangement on the other side. Hence, no. of possible arrangements = 5!/2!3! = 10