How many ways can the letters of the word Intermediate be arranged so i the vowels always occupy even places II the vowels and consonants do not alter?
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In how many ways can be the letter of the word ‘STRANGE’ be arranged so that(a) The vowel may appear in the odd places(b) The vowels are never separated(c) The vowels never come togetherAnswer
Hint:There are 7 letters in the word ‘STRANGE’ in which two are vowels. There are four odd places. So, we have to find the total number of ways for these four places. The other 5 letters may be placed in the five places in 5! ways. By this we get the total number of required arrangements. If vowels are never separated, in this case consider all the vowels a single letter. If all the vowels never come together in this case, we subtract all vowels together from the total number of
arrangements.Complete step-by-step answer: (b) Now, vowels are not to be separated. So, we consider all vowels as a single letter. (c) Now, for the number of arrangements when all vowels never come together, we subtract the total arrangement where all vowels have occurred together from the total number of arrangements. Home » In how many ways can the letters of the word ‘INTERMEDIATE’ be arranged so that: (i) The vowels always occupy even places? (ii) The relative orders of vowels and consonants do not change? In how many ways can the letters of the word ‘INTERMEDIATE’ be arranged so that: (i) The vowels always occupy even places? (ii) The relative orders of vowels and consonants do not change? In how many ways can the letters of the word ‘INTERMEDIATE’ be arranged so that: (i) The vowels always occupy even places? (ii) The relative orders of vowels and consonants do not change? (i) There are 6 even places and 6 vowels out of which 2 are of 1 kind, 3 are of the 2nd kind The vowels can be arranged in = \(\frac{6!}{2!3!}\) = 60 There are 6 consonants out of which 2 is of one kind Number of permutations = \(\frac{6!}{2!}\) = 360 ⇒ Total number of words = 360 x 60 = 21600 (ii) There are 6 vowels to arrange in \(\frac{6!}{2!3!}\) There are 6 consonants which can be arranged in \(\frac{6!}{2!}\) ⇒ Total number of ways = \(\frac{6!}{2!3!}\times\frac{6!}{2!}\) = 21600 How many ways can the letters of the word Intermediate be arranged so that I the vowels always occupy even places?Required number of ways = 21600.
How many words can be made with letters of the word intermediate If the positions of the vowels and consonants are preserved?Total number of words formed =(360×60)=21600.
How many words can be made with letters of the word intermediate if no vowel is between two consonants?The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is 151200.
How many words can intermediate make?The number of words which can be made form letters of the word INTERMEDIATE i 907200 if words start I and end with (B) 21600 if vowels and consonants occupy their originaly places (C) 43200 if vowels and consonants occur alternatively (D) 32400 if all the vowels occur together.
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