SOLUTION: How many arrangements of the word REVISITED are there with vowels not in increasing order and no consecutive E�s and no consecutive I�s? (Normal order is a,e,I,o,u) What i h

Question 32083: How many arrangements of the word REVISITED are there with vowels not in increasing order and no consecutive E�s and no consecutive I�s?
(Normal order is a,e,I,o,u)
What i have done is by direct apporach and i have listed all the cases as prescribed and i got 20 cases in which each case has 5! distrubution of the remaining alphabets in between the specified directions. So the answer i am getting is 5!*20. I am not sure whtehr it is correct.
Thanks in advance.

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
How many arrangements of the word REVISITED are there with vowels not in increasing order and no consecutive E�s and no consecutive I�s?
THERE ARE 9 LETTERS COMPRISING 5 CONSONANTS....R,V,S,T,D.....AND 4 VOWELS ....
2 NOS.I , 2 NOS. E.
(Normal order is a,e,I,o,u)
What i have done is by direct apporach and i have listed all the cases as prescribed and i got 20 cases in which each case has 5! distrubution of the remaining alphabets in between the specified directions. So the answer i am getting is 5!*20. I am not sure whtehr it is correct.
Thanks in advance.
NO CONSECUIVE E'S AND I'S..SO WE HAVE TO INTERPOSE E OR I OR CONSONANT.BUT IF WE INTERPOSE E OR I THEN IT WOULD MEAN VIOLATION OF NO INCREASING ORDER.SO WE HAVE TO INTERPOSE ONLY CONSONANTS.
SO PLACE 5 CONSONANTS FIRST...THESE CAN BE ARRANGED IN 5! WAYS=120 WAY.
NOW THERE ARE 4 INTESPACES BETWEEN THESE 5 CONSONANTS AND 2 OUTER SPACES...6 IN ALL WHERE WE CAN PLACE VOWELS.
LET US CALL THEM 1,2,3,4,5,6 FROM LEFT TO TIGHT.2 NOS.E'S HAVE TO FOLLOW 2 NOS. I'S.SO I'S CAN TAKE UP TO POSITION 4.THAT IS THEY CAN BE IN
12,13,14,23,24,34 POSITIONS ONLY.CORRESPONDINGLY..
12.......E'S CN BE PLACED IN 4 PLACES IN 4!/(2!*2!)=6 WAYS.
13.......E'S CN BE PLACED IN 3 PLACES IN 3!/(2!*1!)=3 WAYS.
14.......E'S CN BE PLACED IN 2 PLACES IN 2!/(2!*0!)=1 WAYS.
23.......E'S CN BE PLACED IN 3 PLACES IN 3!/(2!*1!)=3 WAYS.
24.......E'S CN BE PLACED IN 2 PLACES IN 2!/(2!*0!)=1 WAYS.
34.......E'S CN BE PLACED IN 2 PLACES IN 2!/(2!*0!)=1 WAYS.
TOTAL..............................................15 WAYS.
SO TOTAL POSSIBLE ARRANGEMENTS = 120*15=1800 WAYS.
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