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Originally Posted by GauravShah Quote: |
Originally Posted by ritika_j In how many ways can the letter of the word "daughter" be arranged such that no vowels are together?
answer this question with explanation | Not sure if my approach is correct still here is my solution.....
Lets place this first V-C-V-C-V...here three vowels can be arrange in 3! = 6 ways and consonants in 5P2 = 5.4 = 20 ways...
hence total = 6*20 = 120 ways...
Now consider the spaces between these that is...
_ - V - _ - C - _ - V - _ - C - _ - V - _
In this stepp we fill these spces will the remaining consonants i.e. 3
This can be done in 6C3 * 3! ways = 20*6 = 120 ways
Hence total no. of ways = 120*120 = 14400 ways If I missed something please reply...
Gaurav. |
well champ, u did miss something..u see "daughter" has 3 vowels nd 5 consonants..so its not necessary to have a vc-vc or -v-c sort of repetitive relatonship..actually the solution is much simpler
step1: calculate all possible permutations of "daughter"= 8!
step2: calculate all cases in which three vowels are always together=3!x6!
step3: calculate cases of two vowels being together..this is slightly tricky, so a diagram follows
_ _ _ _ _ _ _ _
1 2 3 4 5 6 7 8 lets consider two vowels at 1-2 position..3C2=3 arrangements
=> rest arrangements can be made in 6! ways
out of these 6! arrangements 5! arrangements will have a vowel at place 3
so we have 3x(6!-5!) ways
now this 1-2 group can be taken in 7 ways
so arrangements for two vowels together are= 7x3x(6!-5!)
hence, we have total arrangements= 8!- [3!x6! + 7x3x(6!-5!)]
