The number of words (with or without meaning) that can be formed from all the letters of the word 'LETTER' in which vowels never come together is_____.

The given word is ‘LETTER’.  We first note the individual letters and how many times each appears:

$$L : 1,\; E : 2,\; T : 2,\; R : 1.$$

We want to count all possible arrangements (words) of these six letters in which the two vowels (both $$E$$’s) are never next to each other.

We begin by finding the total number of arrangements of all six letters without any restriction. The general formula for permutations of $$n$$ objects where some objects repeat is stated first:

$$\text{If } n_1, n_2, \ldots, n_k$$ are the repetition counts, then $$\dfrac{n!}{n_1!\,n_2!\,\ldots\,n_k!}.$$

Here, out of the six letters, the letter $$E$$ repeats $$2$$ times and the letter $$T$$ repeats $$2$$ times. Therefore,

Total unrestricted arrangements $$= \dfrac{6!}{2!\,2!}.$$

Using the factorial definition $$n! = n \times (n-1) \times \cdots \times 1,$$ we have

$$6! = 720,\qquad 2! = 2.$$

Substituting, we get

$$\dfrac{6!}{2!\,2!}= \dfrac{720}{2 \times 2}= \dfrac{720}{4}=180.$$

Now we must subtract those arrangements in which the two vowels come together. To enforce “togetherness,” we tie the two $$E$$’s into a single super-letter $$\bigl(EE\bigr).$$ Treating this super-letter as one entity, the objects we must arrange are

$$(EE),\,L,\,T,\,T,\,R.$$

That is a total of $$5$$ objects, with $$T$$ repeating twice. Using the same repetition formula, the number of arrangements with the two vowels side by side is

$$\dfrac{5!}{2!}.$$

Again, evaluate:

$$5! = 120,\qquad 2! = 2,$$

so

$$\dfrac{5!}{2!}= \dfrac{120}{2}=60.$$

Finally, to get the arrangements in which the vowels are never together, we subtract the “together” count from the total:

$$\text{Required number}= 180 - 60 = 120.$$

So, the answer is $$120$$.