How many arrangements of MATHEMATICS are there in which each consonant is adjacent to a vowel? How many arrangements of MATHEMATICS are there in which each consonant is adjacent to a vowel? 
Now I have calculated it to be 
$$
11 \cdot \frac{7!}{2!2!} \cdot \frac{4!}{2!}
$$
for 11 distinct distributions of consonants among vowels with repetitions making sure that no two vowels are together. 
(If not, then at least one consonant will be not adjacent to any of the four vowels.)
I'm not sure about it and I would like to verify my answer. Any help will be appreciated.
 A: With eight consonants there's only one way to distribute consonants and vowels (cvccvccvccvc) and removing one of them can only be done in five distinct ways (with vowels denoted by - for better visual appearance):

*

*-cc-cc-cc-c


*c-c-cc-cc-c


*c-cc-c-cc-c


*c-cc-cc-c-c


*c-cc-cc-cc-
Then it's only to place the consonants in ${7!\over2!2!}$ ways (since the order of m's and t's doesn't matter) and vowels in ${4!\over2!}$ ways (since the order of a's doesn't matter).
The answer is: $5{7!\over2!2!}{4!\over2!}$
A: You can visualize each vowel as having two arms to which consonants can be attached
$-V-\;\;-V-\;\;-V-\;\;-V-$
There are 8 such arms,and for placing 7 consonants, 1 arm to be left, but if you leave out an arm between two vowels, one of the arms "disappears", so in effect there are only 5 ways of leaving out an arm,
$$\text{ thus answer} = 5 \cdot \frac{7!}{2!2!} \cdot \frac{4!}{2!}$$
ps
If $c$ is the # of consonants, and $v$ the # of vowels, the coefficient of 5 can be obtained as
$$\sum_{k=c-v}^{v-1}\binom{v-1}k\binom{v+1-k}{c-2k}$$
