A word of at least `5` letters is made at random from `3` vowels and `3` constants, all the letters being different. The probability that no consonant falls between any two vowels in the word is

Options

A. `(9)/(20)`
B. `(9)/(10)`
C. `(7)/(10)`
D. `(11)/(20)`

Answer

D

Explanation

`(d)` `E_(5)=` the event of making words of `5` letters
`E_(6)=` the event of making words of `6` letters
`P(E_(6))=(4!xx3!)/(6!)`, arrangement being done by considering all the vowels together for calculating `n(E_(6))`
`P(E_(5))=("^(3)C_(2)*4!2!+^(3)C_(2)*3!3!)/('^(6)P_(5))` {`:'` there are `2` vowels , `3` constants or `3` vowels, `2` constants}
`:.` The required probability
`=P(E_(6))+P(E_(5))=(4!*3!+3*4!*2!+3*3!*3!)/(6!)`
`=(144+144+108)/(720)=(396)/(720)=(11)/(20)`