Arranging JUDGE with vowels together: How many permutations of J, U, D, G, E place the two vowels together?
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A48
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B24
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C120
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D60
Answer
Correct Answer: 48
Explanation
Introduction / Context:We want permutations of the letters of JUDGE with the vowels U and E adjacent. This is a classic “keep-a-group-together” arrangement question.
Given Data / Assumptions:Letters: J, U, D, G, E (all distinct). Vowels: U, E; consonants: J, D, G.
Concept / Approach:Treat the two vowels as a single block, then permute that block with the three consonants. Inside the block, the two vowels can swap.
Step-by-Step Solution:
Entities to arrange: [UE], J, D, G ⇒ 4 distinct entitiesArrange entities: 4! = 24Internal order of vowels: 2! = 2Total = 24 * 2 = 48Verification / Alternative check:Complementary counting (total 5! minus permutations with vowels separated) also yields 48 after careful inclusion–exclusion.
Why Other Options Are Wrong:24 misses the internal swap; 60 or 120 overcount beyond constraints.
Common Pitfalls:Forgetting the 2! within the vowel block.
Final Answer:48