Question

There are 20 seats in some rows of a movie theatre and 5 people sit randomly in this row. The probability that no two persons sit side by side, is
$$\begin{gathered} (a){\displaystyle \frac{91}{323}} \\ (b){\displaystyle \frac{232}{323}} \\ (c){\displaystyle \frac{120}{323}} \\ (d)\text{ none of these} \end{gathered}$$

Answer 1

Hint: In this question first find the total possible outcomes by making 5 people sit in 20 seats of a row. Then the favorable number of outcomes are computed by considering the group of 5 people as 1 subtracted by the total possible cases.

Complete Step-by-Step solution:
Total number of seats in a row is 20.
Total number of persons is 5.
Now the number of ways to arrange 5 persons on 20 seats is ${}^{20}{P}_5$.
Now consider 5 persons as 1 so the total number of seats become (20 – 5 + 1) = 16
So the number of ways to arrange 1 person (which is equal to 5) on 16 seats is${}^{16}{P}_5$.
So the number of ways so that two persons sit side by side is =${}^{16}{P}_5$.
So the number of ways so that two persons sit side by side is = ${}^{20}{P}_5-{}^{16}{P}_5$
Now as we know that the probability (P) is the ratio of favorable number of outcomes to total number of outcomes that is $\text{P = }{\displaystyle \frac{\text{Favorable outcome}}{\text{Total number of outcome}}}$
$\Rightarrow P={\displaystyle \frac{{}^{20}{P}_5-{}^{16}{P}_5}{{}^{20}{P}_5}}$
Now as we know that $${}^n{P}_r={\displaystyle \frac{n!}{\left(n-r\right)!}}$$ so use this property in above equation we have,
$\Rightarrow P={\displaystyle \frac{{\displaystyle \frac{20!}{\left(20-5\right)!}}-{\displaystyle \frac{16!}{\left(16-5\right)!}}}{{\displaystyle \frac{20!}{\left(20-5\right)!}}}}$
Now simplify this we have,
$\Rightarrow P={\displaystyle \frac{{\displaystyle \frac{20!}{15!}}-{\displaystyle \frac{16!}{11!}}}{{\displaystyle \frac{20!}{15!}}}}=1-{\displaystyle \frac{16!\times 15!}{20!\times 11!}}$
$$\begin{gathered} \Rightarrow P={\displaystyle \frac{16!\times 15!}{20!\times 11!}}=1-{\displaystyle \frac{16!\times 15\times 14\times 13\times 12\times 11!}{20\times 19\times 18\times 17\times 16!\times 11!}}=1-{\displaystyle \frac{15\times 14\times 13\times 12}{20\times 19\times 18\times 17}}=1-{\displaystyle \frac{91}{323}} \\ \Rightarrow P={\displaystyle \frac{232}{323}} \end{gathered}$$
So this is the required answer.
Hence option (B) is correct.

Note: The concept behind taking group of 5 people into one was to find the probability of no two of them sitting together, thus instead of finding it directly we have found it conversely, if we make them sit together and then subtract from the total possible arrangements than it gives the arrangements of them no being seated together. This concept helps solve problems of this kind.