The probability that in a random arrangement of the letters of the word “FAVOURABLE”, the two ‘A’ do not come together is

Question

The probability that in a random arrangement of the letters of the word “FAVOURABLE”, the two ‘A’ do not come together is (A) (1/5) (B) (1/10) (C) (9/10) (D) (4/5)

Tardigrade Admin · Accepted Answer

Total number of arrangements of the letters of the word

“ F A V O U R A B L E

” is

\frac{10 !}{2 !}

Now, say

2 A

’s as one letter

∴

Total number of arrangement if ‘

AA

’ consider one letter is

9!

∴

Number of ways in which

‘ AA

’ never come together is

= \frac{10 !}{2 !} - 9!

∴

Required probability

= \frac{\frac{10 !}{2 !} - 9 !}{\frac{10 !}{2 !}} = \frac{4}{5}

Q. The probability that in a random arrangement of the letters of the word $“ F A V O U R A B L E$ ”, the two $‘ A ’$ do not come together is

$\frac{1}{5}$

$\frac{1}{10}$

$\frac{9}{10}$

$\frac{4}{5}$

Q. The probability that in a random arrangement of the letters of the word “FAVOURABLE”, the two ‘A’ do not come together is

51​

101​

109​

54​

Q. The probability that in a random arrangement of the letters of the word $“ F A V O U R A B L E$ ”, the two $‘ A ’$ do not come together is

$\frac{1}{5}$

$\frac{1}{10}$

$\frac{9}{10}$

$\frac{4}{5}$