The number of permutations of the characters in LILAC so that no character appears in its original position, if the two L’s are indistinguishable, is _________.

Let the positions be indexed $1, 2, 3, 4, 5$ corresponding to the original string $\text{L}_1\text{I}_2\text{L}_3\text{A}_4\text{C}_5$.

The letters are two $\text{L}$'s, one $\text{I}$, one $\text{A}$, and one $\text{C}$. In the original word:

• Position 1 has $\text{L}$

• Position 2 has $\text{I}$

• Position 3 has $\text{L}$

• Position 4 has $\text{A}$

• Position 5 has $\text{C}$

For no character to appear in its original position:

1. The two $\text{L}$'s cannot be placed at positions 1 or 3. Thus, the two $\text{L}$'s must be placed in a subset of size 2 from positions $\{2, 4, 5\}$. The number of ways to choose 2 positions for the $\text{L}$'s is $\binom{3}{2} = 3$ ways:

   • Case 1: $\text{L}$'s at positions $\{2, 4\}$ Remaining positions to fill are $\{1, 3, 5\}$ with characters $\{\text{I}, \text{A}, \text{C}\}$.

     • Position 1 cannot have $\text{I}$? No, position 1 originally had $\text{L}$, so any character can go there.

     • Position 3 originally had $\text{L}$, so any of $\{\text{I}, \text{A}, \text{C}\}$ can go there.

     • Position 5 originally had $\text{C}$, so position 5 cannot have $\text{C}$.

     • Total permutations of $\{\text{I}, \text{A}, \text{C}\}$ in positions $\{1, 3, 5\}$ is $3! = 6$.

     • Permutations where $\text{C}$ is at position 5 is $2! = 2$.

     • Valid assignments for this case = $6 - 2 = 4$.

   • Case 2: $\text{L}$'s at positions $\{2, 5\}$ Remaining positions are $\{1, 3, 4\}$ with characters $\{\text{I}, \text{A}, \text{C}\}$.

     • Position 4 originally had $\text{A}$, so position 4 cannot have $\text{A}$.

     • Positions 1 and 3 have no restriction on $\{\text{I}, \text{A}, \text{C}\}$.

     • Valid assignments = $3! - 2! = 6 - 2 = 4$.

   • Case 3: $\text{L}$'s at positions $\{4, 5\}$ Remaining positions are $\{1, 2, 3\}$ with characters $\{\text{I}, \text{A}, \text{C}\}$.

     • Position 2 cannot have $\text{I}$.

     • Positions 1 and 3 have no restriction.

     • Valid assignments = $3! - 2! = 6 - 2 = 4$.

Summing all cases:

Thus, the correct answer is 12.