We have, 6 objects {A}, {P}, {U}, {R}, {B}, {A} and there are 2 A's, so, there are 3 vowels (2 {A}’s and 1 {U}) and 3 consonants ({P}, {R}, {B}).

The word arrangement where, vowels and consonants are alternate is shown below:

Arrangement-1:

Arrangement-2:

The vowel positions can be filled in ways, and the consonant positions can be filled in 3! ways.

As, the operations are dependent so, the number of ways to fill the 6 positions

So, the number of words that can be made by re-arranging the letters of the word APURBA so that vowels and consonants are alternate is

[as, we have 2 possible arrangements]

= 36.