Permutation in PERMUTATIONS

Question

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.

Answer 1

$$PERMUTATIONS$$ 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}$

Answer 2

$PERMUTATION=A^1E^1I^1M^1N^1O^1P^1R^1S^1T^2U^1=C^7V^5$

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