The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is _________.

We first look at the word ‘VOWELS’. It has $$6$$ distinct letters: $$V,\,O,\,W,\,E,\,L,\,S$$. Out of these, $$O$$ and $$E$$ are vowels, while $$V,\,W,\,L,\,S$$ are consonants.

We wish to count all possible six-letter arrangements that can be made with these letters, but in such a way that the four consonants never come together in one single block.

To obtain this count, it is convenient to use the basic counting principle

Total required number $$=\ ($$ Total permutations of all six letters $$)\;-\;($$ Permutations in which all consonants are together $$).$$

We now evaluate each term separately.

1. The formula for the number of permutations of $$n$$ distinct objects is $$n!$$ (“$$n$$ factorial”). Here $$n=6$$, so

Total permutations of all six letters $$=6!$$ $$=6\times5\times4\times3\times2\times1$$ $$=720.$$

2. Next, we count the unwanted arrangements in which all four consonants $$V,\,W,\,L,\,S$$ stand side by side. Treat these four consonants as a single super-letter or block. Together with the remaining two vowels $$O$$ and $$E$$, we now have only $$3$$ “letters” to arrange:

[VWLS] , O , E.

The number of ways to arrange $$3$$ distinct objects is $$3!$$. Inside the block [VWLS] the four consonants themselves can be permuted in $$4!$$ ways. By the multiplication principle, the total number of such “all-consonants-together” words is

$$3!\times4!$$ $$=(3\times2\times1)\times(4\times3\times2\times1)$$ $$=6\times24$$ $$=144.$$

3. Finally, we subtract these unwanted cases from the total:

Required number $$=720-144$$ $$=576.$$

So, the answer is $$576$$.