How many different ways can the letters of the word judge be rearranged such that the vowels always come together?
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Description for Correct answer: 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. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Solution The correct option is A 48The 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 = 48How many ways can the word judge be arranged so that the vowels always come together?= 48. Q. In how many different ways can the letters of the word CHASE be arranged such that the vowels always come together. Q.
How many different ways can the letters of the word position be arranged so that the vowels always come together?Since, the vowels have to be together, we can say that we have to arrange the groups (C), (R), (P), (R), (T), (N) and (OOAIO) among themselves. Considering the objects of the same type, this can be done in. 2 ! = 2520 ways.
How many combination of letters can be formed using all the letters of the word fudge so that the vowels are always together?Required number of ways = (120 x 6) = 720.
How many of them have all vowels occur together?Total numbers of letters, we have to arrange = 5+1=6. Total number of arrangements when all vowels occur together=720×6=4320.
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