Permutation in PERMUTATIONS In the word $PERMUTATIONS$. What is the number of permutations so that a vowel must be between two consonants and a letter can be used only once.
 A: $$PERMUTATIONS$$
Consonant:$PRMTTNS$
Vowel:$EUAIO$
$$ \boxed{consonant} \boxed{vowel} \boxed{consonant} $$
$Case:1$
If the consonants are different $6P2 * 5P1= 150$
$case:2$
If the consonants are same $1 * 5P1= 5$
Total number of permutations $150 + 5 =$ $ \boxed {155}$
A: $PERMUTATION=A^1E^1I^1M^1N^1O^1P^1R^1S^1T^2U^1=C^7V^5$ 
where $C$ means consonant and $V$ vowel
First let us count the words with five letters $V$ and  7 letters $C$ which have the subword $CVC$, this is the same as the total number of words whith five $V$'s and $7$ C's minus the words that don't have $CVC$ as a subword.
How many letters don't have $CVC$ as a subword? 
Classify according to how the letter's $V$ are seperated:
$VV-VVV$ by stars and bars we need to divide the $7$ letter $C$'s into $3$ sections, thus there are $\binom{9}{2}=36$
$VVV-VV$ same as before, $36$ words
$VVVVV$ we need to divide the $7$ letters into the right and left, there are $8$ ways to do it.
So there are $80$ words that don't contain $CVC$ as a subword, if we take into account there are $\frac{12!}{5!7!}=792$ words we get there are $712$ words with $7$ C's and $5$ V's that contain $CVC$ as a subword.
Given a word there are $\frac{7!}{2}$ ways to arrange the consonants and $5!$ ways to arrange the vowels, thus there are $\dfrac{712\cdot7!\cdot5!}{2}=215,308,800$ words where there is a vowel surrounded by consonants
A: Your question is equivalent to "How many words can we make from the word PERMUTATIONS in which no two vowels are come together? "
First list out the vowels and consonants.
Vowels: E, U, A, I, O
Consonants: P, R, M, T, T, N, S
Now,
P_R_M_T_T_N_S
If you filling the five vowels in any blank space yiedls a words which having no two vowels together and each vowel is sorrounded by consonants.
Let's start a computation.
There are 7!/2! Words like above.
And there are 6 blank spaces for filling 5 consonants. So that there are 6C5=6 ways to complete this job.
And among the 5 vowels we can shuffle these in 5! Ways.
Thus finally we have 6×5!×(7!/2!)= 1,814,400 words. 
