# How many different words can be formed of the letters of the word MALENKOV so that no two vowels are together?

Feb 12, 2016

None

#### Explanation:

I am assuming you mean using all 8 letters and creating actual words not gibberish. There are no 8 letter words in the English language that use those letters.

Mar 4, 2016

$14400$

#### Explanation:

Start by looking at the pattern of vowels and consonants, ignoring anything more specific.

There are three vowels, which will partition an arrangement of letters into four unbroken sequences (possibly empty) of consonants: The first being before the first vowel, the second being between the first and second vowels, etc.

In order that no two vowels are adjacent, each of the two middle sequences must contain at least one consonant. So allocate one consonant to each of the two middle sequences.

That leaves us with 3 consonants to share between 4 sequences.

That can be done in the following ways:

$3 + 0 + 0 + 0$
$0 + 3 + 0 + 0$
$0 + 0 + 3 + 0$
$0 + 0 + 0 + 3$
(i.e. $4$ cyclic variants)

$2 + 1 + 0 + 0$
$2 + 0 + 1 + 0$
$2 + 0 + 0 + 1$
$\times 4$ for cyclic variants

$1 + 1 + 1 + 0$
$\times 4$ for cyclic variants

That is, a total of $20$ different ways.

Then teasing apart the subsequences of vowels and consonants, there are 3! xx 5! possible orderings for each of these $20$ partition schemes.

So the overall total number of possible arrangements satisfying the condition is:

20 xx 3! xx 5! = 20 xx 6 xx 120 = 14400