Question:

**Step 1:** For an arrangement of 11 characters to read the same forwards and backwards, it must be symmetric about the center position (position 6). **Step 2:** We have 5 A's and 6 B's. For a palindrome of odd length, the center position must be occupied by a character with an odd total count. Since there are 5 A's (an odd count) and 6 B's (an even count), the center position (6th) must be an A. **Step 3:** After placing one A at the center, 4 A's and 6 B's remain. These 10 characters must be arranged in the remaining 10 positions (1-5 and 7-11) such that positions 7-11 mirror positions 1-5. This means we only need to determine the arrangement for the first 5 positions (1-5). **Step 4:** Let $x$ be the number of A's placed in the first 5 positions and $y$ be the number of B's placed in the first 5 positions. By symmetry, $x$ A's will also be in positions 7-11, and $y$ B's will be in positions 7-11. The total number of A's used is $1$ (for the center) $+ 2x$. Since there are 5 A's in total, we have: $$ 1 + 2x = 5 \implies 2x = 4 \implies x = 2 $$ The total number of B's used is $2y$. Since there are 6 B's in total, we have: $$ 2y = 6 \implies y = 3 $$ Thus, in the first 5 positions, we must arrange 2 A's and 3 B's. **Step 5:** The number of ways to arrange 2 A's and 3 B's in 5 distinct positions is given by the multinomial coefficient: $$ \frac{5!}{2!3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)} = \frac{120}{2 \times 6} = \frac{120}{12} = 10 $$ Alternatively, this can be calculated as the number of ways to choose 2 positions for the A's out of 5 available positions: $$ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = 10 $$ Therefore, there are 10 ways to arrange 5 A's and 6 B's in a row such that the arrangement reads the same backwards and forwards.