Question

The letters of the word "RANDOM" are arranged in all possible ways. The number of arrangements in which there are $2$ letters between 'R' and 'D' is

• A. $36$
• B. $48$
• C. $72$
• D. $144$

Answer 1

The correct option is D $144$

Since there are $6$ letters and there should be $2$ letters between $R$ and $D$.
$\therefore R$ and $D$ can be at
$\left(1\right)$ $1,4$ position
$\left(2\right)$ $2,5$ position
$\left(3\right)$ $3,6$ position
Now the remaing letters can be arranged in $4!$ ways, also $R$ and $D$ can be interchanged bwtween themselves in $2!$ ways.
$\therefore$ Number of required arrangements will be $=2!\times 4!+2!\times 4!+2!\times 4!=3\times 2!\times 4!=144$