Word-formation count: Using each letter only once, how many meaningful English words can be formed from the letters L, E, G, U?

Difficulty: Easy

Correct Answer: Two

Explanation:


Introduction / Context:
This question tests your ability to form valid English words from a fixed multiset of letters, with the constraint that each letter may be used at most once per word. The letter bag is L, E, G, U (one of each). The task is to count how many distinct, common-dictionary words you can make under these rules.


Given Data / Assumptions:

  • Available letters: {L, E, G, U}.
  • Each letter can be used once per word.
  • We count standard English words that appear in common dictionaries (no proper nouns, abbreviations, or rare technical forms).


Concept / Approach:
Enumerate plausible permutations that could yield everyday words. A practical strategy is to consider likely onsets (GL-, L-, G-) and check common endings (-UE, -UE, -UG, etc.). Two familiar words emerge quickly: GLUE and LUGE. Both use all four letters exactly once and are common in everyday English: “glue” (noun/verb) and “luge” (winter sliding sport). No other common permutations (e.g., GULE, EGUL, ULEG) yield standard words in general-purpose usage.


Step-by-Step Solution:
1) List permutations with GL-: GLUE ✓.2) List permutations with L-: LUGE ✓.3) Reject non-words like GULE (archaic/rare), EGUL, ULEG.4) Count valid everyday words: 2.


Verification / Alternative check:
Cross-check in a standard learner’s dictionary: both “glue” and “luge” are attested, common entries. Other sequences do not appear as standalone everyday words, ensuring the count remains two.


Why Other Options Are Wrong:

  • None / One: Under-count; both GLUE and LUGE are valid.
  • Three: Over-count; no third common word from these exact letters.


Common Pitfalls:
Including rare or archaic forms, proper nouns, or misspellings. The question intends everyday vocabulary, where exactly two words fit.


Final Answer:
Two

More Questions from Alphabet Test

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion