Question

A car is parked among N cars standing in a row, but at either end. On his return, the owner finds that exactly r of the N places is still occupied. The probability that both the places neighbouring his car are empty is
(a) $\dfrac{\left( r-1 \right)!}{\left( N-1 \right)!}$
(b) $\dfrac{\left( n-1 \right)!\left( N-r \right)!}{\left( N-1 \right)!}$
(c) $\dfrac{\left( N-r \right)\left( N-r-1 \right)}{\left( N-1 \right)\left( N-2 \right)}$
(d) $\dfrac{^{N-r}{{C}_{2}}}{^{N-1}{{C}_{2}}}$

Answer 1

- Hint: First of all, we will find the number of possibilities of the cars being arranged in the parking spaces. Then we will find the number of ways the cars occupy the parking space such that the owner’s car doesn’t have any car on either side. Then, to find the probability, we find the quotient of the number of possibilities with no car on either side of the owner's car and the number of ways the cars are arranged in parking space.


Complete step-by-step solution -
Total number of spaces for car parking are N.
The driver parks car in one of the spaces other than the ends. Yet, we do not know anything about the other cars in the parking lot.
On the owner’s return, he finds out that r spaces are filled. Therefore, there are (r – 1) cars other than his own car.
And the total number of spaces available other than the space occupied the owner’s car is (N – 1).
Number of ways of choosing r items from N available items is given by ${}^{N}{{C}_{r}}$, were ${}^{N}{{C}_{r}}=\dfrac{N!}{r!\left( N-r \right)!}$
So, the number of possibilities of (r – 1) cars occupying spaces out of the (N – 1) available spaces is given by ${}^{N-1}{{C}_{r-1}}$.
Now, we will find the number of possibilities in which there are no cars on either side of the owner’s car. The owner’s car occupies 1 place and 2 spaces on either side of the owner’s car cannot be occupied. Thus, the total number of spaces on which the remaining cars are parked is (N – 3).
And the number of cars that can be arranged in these spaces except the owner’s car is (r – 1).
Thus, the total number of possibilities of (r – 1) cars occupying spaces out of the (N – 3) available spaces is given by ${}^{N-3}{{C}_{r-1}}$.
Therefore, the probability that both the places neighbouring his car are empty is given by quotient of number of possibilities of not having any cars on either side of the owner’s car and number of possibilities of arranging (r – 1) cars in (N – 1) cars.
\[\begin{align}
  & \Rightarrow P=\dfrac{^{N-3}{{C}_{r-1}}}{^{N-1}{{C}_{r-1}}} \\
 & \Rightarrow P=\dfrac{\dfrac{\left( N-3 \right)!}{\left( r-1 \right)!\left( N-3-\left( r-1 \right) \right)!}}{\dfrac{\left( N-1 \right)!}{\left( r-1 \right)!\left( N-1-\left( r-1 \right) \right)!}} \\
 & \Rightarrow P=\dfrac{\left( N-3 \right)!\left( N-r \right)!}{\left( N-1 \right)!\left( N-r-2 \right)!} \\
 & \Rightarrow P=\dfrac{\left( N-3 \right)!\left( N-r \right)\left( N-r-1 \right)\left( N-r-2 \right)!}{\left( N-1 \right)\left( N-2 \right)\left( N-3 \right)!\left( N-r-2 \right)!} \\
 & \Rightarrow P=\dfrac{\left( N-r \right)\left( N-r-1 \right)}{\left( N-1 \right)\left( N-2 \right)} \\
\end{align}\]
Hence, option (c) is the correct option.

Note: It is to keep in mind that the position of the owner’s car doesn’t change in any scenario. It occupies the same place throughout. The place of the other cars changes in each different possibility. N! is defined as N(N – 1)(N – 2)….1.