Question

There are $20$ seats in some row of a movie theatre and $5$ persons sit randomly in this row. The probability that no two persons sit side by side, is

• A. $\frac{91}{323}$
• B. $\frac{232}{323}$
• C. $\frac{120}{323}$
• D. None of these

Answer 1

The correct option is A $\frac{91}{323}$

Total ways ${=}^{20}{P}_5$
Let $x_0=$ Number of sects to the left of the first person from the left.$x_i(1\le i\le 4),$
the number of empty seats between the $i^{th}$ and ${(i+1)}^{th}$ person from left,
and $x_5$ the number of emply seats on the last person.
$x_0\ge 0,x_5\ge 0$ and $x_i\ge 1$ for $1\le i\le 4$
$x_0+x_1+...+x_5=15$
$=$ coefficient of $x^5$ in ${(1+x+x^2+...)}^2{(x+x^2+...)}^4$
${=}^{20-5+1}{C}_5{=}^{16}{C}_5$
Required probability $\frac{{}^{16}{C}_5.5!}{{}^{20}{P}_5}=\frac{{}^{16}{P}_5}{{}^{20}{P}_5}=\frac{91}{323}$