Question

The number of ways in which the letters of the word ${}^{\prime }ARRANG{E}^{\prime }$ can be arranged so that the two $A$'s are together but not two $R$'s is

• A. $120$
• B. $240$
• C. $300$
• D. $360$

Answer 1

The correct option is B $240$

The letters of word $ARRANGE$ has $2A$'s and $2R$’s, i.e.. total $7$ letters.
Taking both $A$ together,
The number of words except $R$
$=4!$
Now, using gap method, choosing $2$ places for $R$ from the gap of $5$
$={}^5{C}_2$
Therefore, the total number of words
$=4!\times {}^5{C}_2=240$