How many words can one form by shuffling the letters of the word QUEUE, if Q is always followed by U? The words thus formed need not necessarily have any meaning.

The correct answer is Option D (12).

Why the correct option is correct: This problem is solved using the standard 'Block Method' (or Tie Method) in combinatorics. The word QUEUE consists of 5 letters: Q, U, U, E, E. The condition mandates that 'Q' is always followed by 'U'. To ensure this, we tie one 'Q' and one 'U' together into a single indivisible block: (QU).

After forming this block, the remaining letters available for arrangement are U, E, and E. We now arrange these four separate entities:

1. The block (QU)
2. The free letter U
3. The letter E
4. The letter E

We have n = 4 total entities. Among them, the letter 'E' is repeated twice (p = 2). (The 'U' inside the block is fixed and distinct from the free 'U'). According to the permutation formula for repeated identical items (n!/p!), the total number of unique words is: 4!/2! = 24/2 = 12

Thus, exactly 12 words can be formed.

Why the wrong options are incorrect:

• Option A (6): This represents 3! = 6. It is the erroneous result of dropping the second 'U' altogether or arranging only the non-block letters without placing the block.
• Option B (8): This is a mathematical distraction, likely resulting from faulty addition (e.g., 4 + 4) instead of applying the fundamental principle of counting.
• Option C (10): This is incorrect and stems from guessing or improperly modifying the unconstrained total permutations (5!/(2! × 2!) = 30) without properly utilizing the block method.

Concluding Takeaway: When a sequence of elements must stay together in a specific order in a permutation problem, apply the Block Method: treat the restricted sequence as a single unit, count the remaining standalone elements to find total entities, and divide the total factorial by the factorials of any identical remaining items.