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# String manipulation

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From a seven letter word by mixing up the letters in the word COMBINE.

a. How may ways can you do this if all the vowels have to be at the beginning?
I dont understand that
Ans. 144

b. How may ways can you do this if no vowel is isolated between two consonants?
I dont get that one either
Ans.1,872

Needs help. Thanks.
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CERTIFIED EXPERT
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Top Expert 2013
Commented:
a. If all the vowels are at the beginning then you have three vowels followed by four consonants.
So P(3,1)*P(4,1) = 3!*4! = 6*24 = 144

CERTIFIED EXPERT
Awarded 2010
Top Expert 2013
Commented:
b. If you know where the consonants and vowels are, then there are 144 combinations with the vowels and consonants in those places (as in a).

So how many different ways can you arrange it so that there are no vowels isolated between two consonants?
There are 13 (I'll let you find them) so 13*144 = 1872
CERTIFIED EXPERT
Most Valuable Expert 2014
Top Expert 2015
Commented:
1,872/144 = 13
13 ways to arrange 3 vowels and 4 consonants such  that no vowel is isolated between two consonants
CERTIFIED EXPERT
Most Valuable Expert 2011
Top Expert 2012
Commented:
There are 13 patters of vowels and consonants where you don't have cvc

ccvvvcc
vccccvv
ccccvvv
cvvvccc
cccvvvc
vcccvvc
cvvcccv
vcvvccc
vvccccv
vvvcccc
vccvvcc
cccvvcv
ccvvccv

within each of those patters there are 3! x 4! permutations of the 3 vowels (3!) and 4 consonants (4!)

so 3! x 4! x 13 = 1872