Arranging JUDGE with vowels together: How many permutations of J, U, D, G, E place the two vowels together?

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:

Verification / 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