Question

A car is parked by an owner in a parking lot of 25 cars in a row, including his car not at either end. On his return he finds that exactly 15 places are still occupied. The probability that both the neighboring places are empty is
A. $\dfrac{{91}}{{276}}$
B. $\dfrac{{15}}{{184}}$
C. $\dfrac{{15}}{{92}}$
D.None of these.

Answer 1

Hint : It is given that on return of the owner, there are 15 places still occupied, which also includes the car of the owner. Since one car out of 15 is the owner’s car, there are 24 places for 14 cars excluding the owner’s car. So, the total number of outcomes is given as $^{24}{C_{14}}$ .

Complete step-by-step answer :
As it is given that when the owner returns then there are still 15 places occupied. So, there are 24 places for 14 cars excluding the owner’s car. Therefore, the total numbers of cars to park the 14 cars in 24 places is given as:
$^{24}{C_{14}}$ = $$ $\dfrac{{24!}}{{14!(24 - 14)!}}$
Now, it is the given condition that the neighboring places are empty, then 14 cars must be parked in,
(25 – (1 owner car) - (2 neighboring places)) = 22 places.
So, the favorable number of ways to park the 14 cars in 22 places is given as $^{22}{C_{14}} = \dfrac{{22!}}{{14!(22 - 14)!}}$
Now, the required probability is calculated by taking the ratio of the favorable number of ways to the total number of ways.
Required probability = $\dfrac{{Favorable\,ways}}{{Total\,ways}}$
Substituting the values of the favorable ways as $^{24}{C_{14}}$ in the above expression.
Required probability
= $\dfrac{{\dfrac{{22!}}{{14!(22 - 14)!}}}}{{\dfrac{{24!}}{{14!(25 - 14)!}}}}$

= $\dfrac{{\dfrac{{22!}}{{14!8!}}}}{{\dfrac{{24!}}{{14!10!}}}}$

Required probability
= $\dfrac{{22!}}{{14!8!}} \times \dfrac{{14!10!}}{{24!}}$
Simplifying the expression by the expansion of factorial.
$\Rightarrow$ Required probability
= $\dfrac{{22!}}{{8!}} \times \dfrac{{10 \times 9 \times 8!}}{{24 \times 23 \times 22!}} \\\ = \dfrac{{10 \times 9}}{{24 \times 23}} \\\ = \dfrac{{90}}{{552}} \\\$
Express the above fraction in the lowest term. So, required probability = $\dfrac{{90}}{{552}}$ = $\dfrac{{15}}{{92}}$
So, the required probability is $\dfrac{{15}}{{92}}$ . It means that the probability that both the neighboring places are empty is $\dfrac{{15}}{{92}}$.
So, the correct answer is “Option C”.

Note : In the solution, we had simplified the value of $^{24}{C_{14}}$ , and the general way to simplify the combination is given as: $^a{C_b}\, = \,\dfrac{{a!}}{{b!(a - b)!}}$
Take care while simplifying the expression by the expression of the factorial. In this question, according to the condition given that two neighboring places are empty, for this we have subtracted 25 – 1 (owner’s car) – 2 (neighboring places) = 22. Then finally we have 22 places to occupy 14 cars. So, the number of cars parked in 22 places is $^{22}{C_{14}}$ .