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# Find the number of words those can be formed by using all letters of the word ' DAUGHTER'. If(i) Vowels occurs in first and last place.(ii) Start with letter G and end with letters H.(iii) Letters G,H,T always occurs together.(iv) No two letters of G,H,T are consecutive(v) No vowel occurs together(vi) Vowels always occupy even place.(vii) Order of vowels remains same.(viii) Relative order of vowels and consonants remains same.(ix) Number of words are possible by selecting 2 vowels and 3 consonants.

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## 'DAUGHTER' :Total words=8Vowels are 'AUE'Consonants are 'DGHTR'(1) Vowels occur in first and last placeFor the first letter, there are 3 possibilitiesso, after filing the first letter, the last letter has 2 possibilitiesFiling the other 6 places, we can do that in 6! ways because all letters are different.So, the number of ways = 3*2*6! =4320 ways(2)Start with letter G and end with letter HRemaining places can be filled in 6! ways=720 ways.(3)Letters G, H, T occurs together alwaysConsider GHT as one objectthen the number of object = 5+1=6Numbers of arranging them=6! waysWe can re-arrange G, H, T among themselves in 3! ways.So, the number of ways = 3!6!=4320 ways.(4)No 2 letters G, H, T are consecutivecase(1): All 3 are consecutiveNumber of ways=4320 wayscase(2): Only G, H are consecutiveConsider GH as one objectthen the number of objects = 6+1=7Numbers of arranging them=7!2! waysBut in these, we also get the cases in which G, H, T come togetherso we need to subtract 4*6!The number of arrangements in which the only G, H are consecutive= 7!2!-4*6!Total number of ways in which at least 2 of G, H, T are consecutive= 4320+3(7!2!-4*6!)=25920Number of ways of arranging such that no two ways in which at least 2 of G, H, T are consecutive=Number of ways-25920=8!-25920=14,400 ways(5) No vowel occur together3 vowels- A, U, E.The same argument as above follows for this.So, number of ways= 14,400(6) Vowels occupy even spaceThere are 4 even places and 3 vowels 'so we can arrange vowels in 43P ways.The remaining 5 letters can be arranged in 5! ways.So, number of ways= 43P×5! =2880 ways,(7) order of vowels remains sameThe total number of arrangements =8! waysBut in these, we cannot change the order of vowels.Suppose the required ways be 'n' multiplied by 3! wayswith 'n' must be given 8!i.e 3!.n=8!n=6720 ways.(8) order of vowels and consonants remains sameHere, there are 5 consonants and 3 vowelsi.e 5!3!n=8!n=56 ways.(9) We need to select two vowels which can be done in 32C ways. Selecting 5 consonants can be done in 53C ways.Rearranging can be done 5! waysSo,number of words that are possible by selecting 2 vowels and 3 consonants = 53C × 32C×5!=3600 ways

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