Question

The number of ways in which the letters of the word ${}^{\prime }ARRANG{E}^{\prime }$ can be arranged so that the two $R$'s are never together is

• A. $360$
• B. $900$
• C. $1260$
• D. $1800$

Answer 1

The correct option is B $900$

The letters of word $ARRANGE$ has $2A$'s and $2R$’s, i.e.. total $7$ letters.
Total number of words
$=\frac{7!}{2!2!}=1260$

The number of word in which $2R$’s are together
$=\frac{6!}{2!}=360$
Hence, the required number of words
$=1260-360=900$