Question 52

Word: LILAC (letters: $L, I, L, A, C$).
Original positions:

• Position 1: L
• Position 2: I
• Position 3: L
• Position 4: A
• Position 5: C

Placement condition: No letter may appear in its original position.

• The two L's cannot be at positions 1 or 3.
• Therefore, the two L's must be placed in positions chosen from $\{2, 4, 5\}$.
• Number of ways to choose positions for the two L's is $\binom{3}{2} = 3$.

Analyze the three cases for remaining letters $\{I, A, C\}$:

1. L's at positions 2 and 4:
   • Remaining positions: $\{1, 3, 5\}$.
   • Forbidden positions for remaining letters: C cannot be at 5 (its original position). I and A have no forbidden positions in $\{1, 3, 5\}$.
   • Total permutations of 3 letters = $3! = 6$. Permutations with C at position 5 = $2! = 2$.
   • Valid arrangements = $6 - 2 = 4$.
2. L's at positions 2 and 5:
   • Remaining positions: $\{1, 3, 4\}$.
   • Forbidden position: A cannot be at position 4.
   • Valid arrangements = $3! - 2! = 6 - 2 = 4$.
3. L's at positions 4 and 5:
   • Remaining positions: $\{1, 2, 3\}$.
   • Forbidden position: I cannot be at position 2.
   • Valid arrangements = $3! - 2! = 6 - 2 = 4$.

Total number of valid permutations: