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 option is D - 12.

[as per provisional answerkey]

Solution

The word given is QUEUE. It contains 5 letters: Q, U, E, U, E.
The condition is that Q must always be followed by U. This means the letters 'Q' and 'U' must appear together in the sequence "QU".

To solve this, we treat the block (QU) as a single entity or a single "super-letter".
Now, we list the remaining letters along with this block:
1. (QU) - treated as one unit
2. E
3. U
4. E

We now have 4 units to arrange: {(QU), E, U, E}.
The number of ways to arrange n items where some are identical is given by the formula: n! / (p! * q! ...), where p and q are the frequencies of repeating items.

In our set of 4 units:
- Total units (n) = 4
- The letter 'E' repeats 2 times.
- The units (QU) and 'U' are distinct from each other in this context because the 'U' inside the block is fixed behind 'Q'.

Calculation:
Number of arrangements = $4!/2!$
= $(4\times 3\times 2\times 1)/(2\times 1)$
= $24/2$
= 12

Final Result: There are 12 such words.

Why the other options are incorrect

• Option (a) - 6: This result would occur if one incorrectly assumed there were only 3 units to arrange or applied an extra division by 2! for the 'U's, forgetting that one 'U' is locked inside the (QU) block and cannot be swapped with the independent 'U' without breaking the "Q followed by U" rule.
• Option (b) - 8: This value does not correspond to any standard permutation calculation for this set of letters and likely results from a counting error.
• Option (c) - 10: This is a common distractor for those who might try to manually list combinations but miss specific permutations involving the placement of the repeating 'E's.

Key Concept

The "String Method" in Permutations: treating a fixed group of letters as a single unit and then applying the formula for permutations of a multiset.