Thirty two persons $X_{1}, X_{2}, \ldots, X_{32}$ are randomly seated around a circular table at equal intervals. Two persons $X_{i}$ and $X_{j}$ are said to be within earshot of each other if there are at most three persons between them on the minor arc joining $X_{i}$ and $X_{j}$. The probability that $X_{1}$ and $\mathrm{X}_{2}$ are within earshot of each other is, $\left(\right.$ Here $\left.\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}\right)$

• A

  $\frac{\left(\begin{array}{c}32 \\ 2\end{array}\right) 30 !}{8(32 !)}$
• B

  $\frac{\left(\begin{array}{c}32 \\ 2\end{array}\right) 30 !}{4(32 !)}$
• Correct answer
  C

  $\frac{8}{31}$
• D

  $\frac{4}{31}$

Solution

Case - 1 No person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{30 ! \times 2 !}{31 !}=\frac{2}{31}$ Case-2 One person between $X_{1} \& X_{2}$
$\frac{{ }^{30} \mathrm{C}_{1}(29) ! \times 2 !}{31 !}=\frac{2}{31}$ Case-3 When 2 person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{{ }^{30} \mathrm{C}_{2} \times 28 ! \times 2 ! \times 2 !}{31 !}=\frac{2}{31}$ Case-4 When 3 person between $X_{1} \& X_{2}$ $\frac{{ }^{30} \mathrm{C}_{3} \times 27 ! \times 2 ! \times 3 !}{31 !}=\frac{2}{31}$ Total $=\frac{8}{31}$

Hence, the correct answer is option: 'C'.