Question

The number of ways in which 5A's and 6B's can be arranged to form a string of text which reads the same way backward and forward is

• A. 10
• B. 24
• C. 42
• D. 15

Answer 1

The correct option is A

10

String is of 11 alphabets and no. of A’s are odd
$\Rightarrow$ 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 $=\frac{5!}{2!3!}=10$