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Question: A car is parked by an owner amongst \(25\) cars in a row, not at either end. On his return, he finds...

A car is parked by an owner amongst 2525 cars in a row, not at either end. On his return, he finds that exactly 1515 places are still occupied. The probability that both neighboring places are empty is
A19276\dfrac{{19}}{{276}}
B 15284\dfrac{{15}}{{284}}
C1592\dfrac{{15}}{{92}}
D125\dfrac{1}{{25}}

Explanation

Solution

Hint: It is given that on the return of the owner, there are 1515 places still occupied, which also contains the car of the owner. Since, one car out of 1515, is the owner's car, so there are 2424 places for 1414 cars excludes the owner's car. So, the total number of outcomes is given as 24C14^{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 ways to park the 14 cars in 24 places is given as:
24C14=24!14!(24−14)!{}^{24}{C_{14}} = \dfrac{{24!}}{{14!\left( {24 - 14} \right)!}}
Now, it is the given condition that the neighbouring places are empty, then 14 cars must be parked in(25−(1 owner car)−(2neighbouring places))=22\left( {25 - \left( {1{\text{ owner car}}} \right) - \left( {2{\text{neighbouring places}}} \right)} \right) = 22 places. So, the favourable number of ways to park the 14 cars in 22 places is given as:
22C14=22!14!(22−14)!{}^{22}{C_{14}} = \dfrac{{22!}}{{14!\left( {22 - 14} \right)!}}
Now, the required probability is calculated by taking the ratio of the favourable number of ways to the total number of ways.

Required Probability=Favourable ways  Total ways{\text{Required Probability}} = \dfrac{{{\text{Favourable ways}}\;}}{{{\text{Total ways}}}}
Substituting the values of the favourable ways as22C14{}^{22}{C_{14}}and total number of ways as 24C14{}^{24}{C_{14}} in the above expression:
Required Probability=22!14!(22−14)!24!14!(25−14)!{\text{Required Probability}} = \dfrac{{\dfrac{{22!}}{{14!\left( {22 - 14} \right)!}}}}{{\dfrac{{24!}}{{14!\left( {25 - 14} \right)!}}}}
Required Probability=22!14!8!24!14!10!{\text{Required Probability}} = \dfrac{{\dfrac{{22!}}{{14!8!}}}}{{\dfrac{{24!}}{{14!10!}}}}
Required Probability=22!14!8!×14!10!24!{\text{Required Probability}} = \dfrac{{22!}}{{14!8!}} \times \dfrac{{14!10!}}{{24!}}
Simplify the expression by the expansion of factorial.
Required Probability=22!8!×10×9×8!24×23×22!{\text{Required Probability}} = \dfrac{{22!}}{{8!}} \times \dfrac{{10 \times 9 \times 8!}}{{24 \times 23 \times 22!}}
Required Probability=10×924×23{\text{Required Probability}} = \dfrac{{10 \times 9}}{{24 \times 23}}
Required Probability=90552{\text{Required Probability}} = \dfrac{{90}}{{552}}
Express the above fraction in the lowest term:
Required Probability=1592{\text{Required Probability}} = \dfrac{{15}}{{92}}
So, the required probability is 1592\dfrac{{15}}{{92}}. It means that the probability that both the neighbouring places are empty is1592\dfrac{{15}}{{92}}.

Therefore, the option C is correct.

Note: In the solution, we had simplified the value of 24C14^{24}{C_{14}}, and the general way to simplify the combination is given as:
aCb=a!b!(a−b)!^a{C_b} = \dfrac{{a!}}{{b!\left( {a - b} \right)!}}
Take care while simplifying the expression by the expansion of the factorial. Try to eliminate the values of factorial without the expansion, because the calculation after the expansion becomes harder.