Option : (C)
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 \(\frac{3!}{2!}\) 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 3 x 3!
= 3 x 6
= 18
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 = 18 x 2
[as, we have 2 possible arrangements]
= 36.
