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Difficulty: Very hardPermutations

Find the number of distinct arrangements of the letters of the word PARALLEL\text{PARALLEL} such that no two letters ’L’\text{'L'} are adjacent.

Answer: 1200

Answer

1200
To ensure no two letters 'L' are adjacent, we use the gap method. First, arrange the 5 non-L letters (P, A, R, A, E). Because 'A' repeats twice, there are 5! / 2! = 60 distinct arrangements. These 5 letters form 6 available gaps (including the ends). Selecting 3 of these 6 gaps to insert the 3 identical 'L's can be done in C(6, 3) = 20 ways. Multiplying these gives 60 × 20 = 1200 valid arrangements.

Step-by-Step Solution

1
Count the frequency of each letter in the word PARALLEL.
The word has 8 letters: 1 P, 2 A's, 1 R, 3 L's, and 1 E.
Recognizing repeated elements is essential for permutations with duplicates.
2
Arrange the non-restricted letters (P, A, R, A, E).
Number of arrangements = 5! / 2! = 60.
The letter 'A' is repeated twice, so we divide 5! by 2!.
3
Calculate the number of available gaps for placing the 3 'L's so that no two are adjacent.
5 arranged letters create 6 gap positions. Choosing 3 gaps gives C(6, 3) = (6 × 5 × 4) / (3 × 2 × 1) = 20 ways.
Placing at most one 'L' per gap guarantees that no two 'L's will be adjacent.
4
Multiply the number of arrangements of non-L letters by the gap choices.
Total arrangements = 60 × 20 = 1200.
By the fundamental counting principle, total arrangements equal the product of independent choices.

Key Concept

Permutations with repeated elements and non-adjacency constraints using the Gap Method
Estimated Time:1m 30s
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Topics

Permutations

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