Question

If the letters of the word $MAHARASTRA$ are permuted at random, then the probability that the two $R^{\prime }s$ come together and no two $A^{\prime }s$ come together is:

• A. $\frac{5}{42}$
• B. $\frac{4}{7}$
• C. $\frac{1}{42}$
• D. $\frac{5}{7}$

Answer 1

The correct option is C $\frac{1}{42}$

$4$ $A$'s are similar, and $2$ $R$'s are similar.
So, $n\left(S\right)=\frac{10!}{4!2!}$

Considering $2$ $R$'s as $1$ unit, we can arrange $1$ unit & $M,H,S,T$ in $5!$ ways. Then there are $6$ gaps between $1$ unit of $R$'s and $M,H,S,T.$ Now $4$ $As$ can be arranged in $6$ gaps in ${(}^6{\text{C}}_4$) ways.
$\therefore$ No. of favourable cases = $5!\times {(}^6{\text{C}}_4)$ ways.
$\therefore$ Required probability $=\left[\frac{5!{\times }^6{\text{C}}_4}{\left(\frac{10!}{4!.2!}\right)}\right]=\frac{5!\times 6!}{10!}=\frac{120}{\mathrm{10.9.8.7}}=\frac{1}{42}$