AdlaT wrote:

Additional. if the word "different" is not given,

Total 5 letter words \(5*5*21*21*21\).

Actually things get much messier than that:

How many 5-letter words can be formed using the letters of the English alphabet that contain 2 different vowels and 3 different consonants?

is now

How many 5-letter words can be formed using the letters of the English alphabet that contain 2 vowels and 3 consonants?

What you've suggested in \(5*5*21*21*21\) is the number of unique values with set positions for vowels and consonants; VVCCC. You might suggest the way to fix this is just 5!. However, when repetition is allowed within the V and the C, you will get duplicates in your words and have to correct for this.

For example, say you get:

AABBB can only form \(5!/(2!3!)\) = 10 different words.

AEBCD can form 5! = 120 different words.

We have to break it down a bit more:

Vowels: Since there are 5, there are 5*5 = 25 different two vowel combinations. 5 of these are pairs: AA, EE, etc. The other 20 are both different.

Consonants: Since there are 21, there are 21*21*21 = 9261. 21 of these are triplets: BBB, CCC, DDD and 3*20*21 = 1260 that contain doubles. (This can be seen by @@#, @#@, #@@ where @ can be any of the 21 consonants and # is any of the remaining 20 consonants.) There are 21*20*19 = 7980 where all are different.

Returning to the original setup we need to choose 2 vowels and 3 consonants. They could have any of these forms:

(Vowels) (Consonants)

(Both Different)(All Different) = 20*7980

(Both Different)(Two the Same) = 20*1260

(Both Different)(All the Same) = 20*21

(Pair)(All Different) = 5*7980

(Pair)(Two the Same) = 5*1260

(Pair)(All the Same) = 5*21

And for each of these scenarios there is a different amount of unique words that can be created:

(Both Different)(All Different) = 5!

(Both Different)(Two the Same) = 5!/2!

(Both Different)(All the Same) = 5!/3!

(Pair)(All Different) = 5!/2!

(Pair)(Two the Same) = 5!/(2!2!)

(Pair)(All the Same) = 5!/(2!3!)

So you need to take the two distributions above to get the total number of unique words given repetition in choosing vowels/consonants.

This is why you can almost always assume the GMAT has a pretty basic combinatorics problem, because a slight change like this would make the problem very long. =)