It would be easier to count the number of words that have $2$ or fewer vowels, and then subtract this number from the total number of words (which you have already computed).

I assume that by "$3$" or more vowels" you mean $3$ or more *occurrences* of vowels, so in particular a word with 2 e's, 2 i's, and the rest consonants qualifies.

How many words are there with no vowels? Clearly
$$21^7$$
if, as per usual convention, we agree that there are $5$ vowels.

How many words with $1$ vowel? **Where** the vowel occurs can be chosen in $\binom{7}{1}$ ways. For each of these ways, the vowel can be chosen in $5$ ways. And once you have done that, the consonants can be filled in in $21^6$ ways, for a total of
$$\binom{7}{1}(5)(21^6)$$

Finally, how many with $2$ vowels? The *location* of the vowels can be chosen in $\binom{7}{2}$ ways. Once this has been done, the actual vowels can be put into these places in $5^2$ ways. And then you can fill in the consonants in $21^5$ ways, for a total of
$$\binom{7}{2}(5^2)(21^5)$$

Add up the $3$ numbers we have obtained, subtract from $26^7$.

Our argument was a little *indirect.* We could instead find, using the same sort of reasoning, the number of words with $3$ vowels, with $4$ vowels, with $5$, with $6$, with $7$, and add up. This is only a little more work than the indirect approach. But any saving of work is helpful! Also, the indirect approach that was described lets us concentrate on pretty simple situations, the most complicated of which is the $2$ vowel case.

**Remark:** The calculation we have done is closely connected to the *Binomial Distribution*, and if you have already covered this, it may be the point of the exercise. So if you know about the Binomial Distribution, imagine the letters to be chosen at random. Then the number of patterns with $3$ or more vowels is the *probability* of $3$ or more vowels, multiplied by $26^7$.

## Best Answer

If we have exactly one repetition is allowed, we start by choosing our three letters: $\binom{26}{3}$. We then choose a letter to be repeated in $\binom{3}{1} = 3$ ways. Finally, we permute our letters in $4!/2!$ ways by the multinomial distribution. By rule of product, we multiply, to get:

$$\binom{26}{3} \cdot 3 \cdot 4!/2!$$

This is correct thinking. Notice that the words with no repetition are disjoint from words with exactly one repetition. So by rule of sum, you add $26!/(26-4)!$ with the quantity I noted above.