How many words with or without meaning can be formed from the letters of the word daughter if Dau always occur together?

Text Solution

Solution : No. of letters in the word 'DAUGHTER' =8 (i) If there is no restriction, No. of permutations to fill 8 places with given 8 letters `= P_(8) = 8! = 40320`. (ii) In the word 'DAUGHTER'. Vowels = A,U,E and Consonants = D,G,T,H,R `rArr` No. of vowels and consonants are 3 and 5 respectively. `:'` A,U,E are always together `:.` Treating these letters as one letter, the letters are `5 +1 = 6` Now the number of arrangement of 6 letters at 6 places `= .^(6)P_(6) = 6! = 720` and Number of arrangement of letters A,U,E at 3 places `= .^(3)P_(3) = 3! =6`. `:.` Required arrangements `= 720 xx 6 = 4320` (iii) For the words starting with A A will be at first place and no of arrangements of remaining 7 letters at remaining 7 places `=.^(7)P_(7) = 7! = 5040`. (iv) For the words starting with A and ending with R, No. of arrangements of remaining 6 letters at remaining 6 places `=.^(6)P_(6) = 6! = 720`.

Text Solution

Solution : The letters of the word daughter are “d,a,u,g,h,t,e,r”. So, the vowels are ‘a, u, e’ and the consonants are “d,g,h,t,r”. (i)Now, all the vowels should come together, so consider the bundle of vowels as one letter, then total letters will be `6`. So, the number of words formed by these letters will be `6!` but, the vowels can be arranged differently in the bundle, resulting in different words, so we have to consider the arrangements of the `3` vowels. So, the arrangements of vowels will be `3!` Thus, the total number of words formed will be equal to `(6!×3!)=4320` (ii)First arrange `5` consonants in five places in `5!` ways. `6` gaps are created. Out of these `6` gaps, select `3` gaps in `6_(C_3)` ways and then make the vowels permute in those `3` selected places in `3!` ways. This leads `5!×6_(C_3)xx3!`=14400.

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