Correct Answer:
Description for Correct answer:
No. of vowels in the word JUDGE = 2 i.e., U and E.
In such cases we treat the group of two vowels as one entity or one letter because they are supposed to always come together. Thus, the problem reduces to arranging 4 letters i.e. J, D. G and [UE] in 4 vacant places.
No. of ways 4 letters can be arranged in 4 places = \[ \Large 4 \times 3 \times 2 \times 1 \] = 24
But the 2 vowels can be arranged among themselves in 2 different ways by interchanging their position. Hence, each of the above 24 arrangements can be written in 2 ways.
Therefore, Required No. of total arrangements = \[ \Large 24 \times 2 \] = 48
Part of solved Permutation and combination questions and answers : >> Aptitude >> Permutation and combination
A. 48
B. 120
C. 124
D. 160
E. None of these
Solution[By Examveda Team]
The given word contains 5 different letters.
Keeping the vowels UE together, we suppose them as 1 letter.
Then, we have to arrange the letters JDG [UE].
Now, we have to arrange in 4! = 24 ways.
The vowels [UE] can be arranged among themselves in 2 ways.
∴ Required number of ways =
[24 × 2] = 48
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Solution
The correct option is A 48
The word 'JUDGE' has 5 letters in which `JDG' are consonants and `UE' are vowels. On keeping vowels together, we get JDG [UE]
∴ Number of arrangements = 4!
× 2 !
=4×3×2×2
=
48