Question

A man parks his car among n cars standing in a row, his car not being parked at an end. On his return he finds that exactly m of the n cars still there. What is the probability that both the cars parked on the two sides of his car have left?
a) $\dfrac{\left( n-m+1 \right)\left( n+m \right)}{\left( n-1 \right)\left( n-2 \right)}$
b) $\dfrac{\left( n-m-1 \right)\left( n-m \right)}{\left( n-1 \right)\left( n-2 \right)}$
c) $\dfrac{\left( n-m+1 \right)\left( n-m \right)}{\left( n-1 \right)\left( n-2 \right)}$
d) $\dfrac{\left( n-m-1 \right)\left( n+m \right)}{\left( n-1 \right)\left( n-2 \right)}$

Answer 1

Hint: In order to this question, you need to know the permutations and combinations. First you have to find the number of ways the m-1 cars can be arranged in the parking area from the n-1 cars. We take m-1 because we do not have to include the car of the man. If we exclude two places on either side of the car of the man. Then we have to find the number of ways to arrange the m-1 cars from the n-3 cars. Then divide both of them to get the probability.

Complete step by step solution:
First you have to find the number of ways the m-1 cars can be arranged in the parking area from the n-1 cars. We take m-1 and n-1 because we do not have to include the car of the man.
Therefore, two arrange m-1 cars from n-1 cars, we use the formula:
${}^{n-1}{{C}_{m-1}}$--- (1)
Next, we exclude two places on either side of the car of the man. Then we have to find the number of ways to arrange the m-1 cars from the n-3 cars.
Therefore, two arrange m-1 cars from n-3 cars, we use the formula:
${}^{n-3}{{C}_{m-1}}$--- (2)
Now, two get the probability, we divide the 2nd equation with the 1st.
Therefore, we get:
$\Rightarrow P=\dfrac{{}^{n-3}{{C}_{m-1}}}{{}^{n-1}{{C}_{m-1}}}$
$\Rightarrow P=\dfrac{\dfrac{\left( n-3 \right)!}{\left( n-2-m \right)!\left( m-1 \right)!}}{\dfrac{\left( n-1 \right)!}{\left( n-m \right)!\left( m-1 \right)!}}$
But $\left( n-m \right)!=\left( n-m-2 \right)!\left( n-m \right)\left( n-m-1 \right)$. Therefore, we get,
$\Rightarrow P=\dfrac{\left( n-m-1 \right)\left( n-m \right)}{\left( n-1 \right)\left( n-2 \right)}$

So, the correct answer is “Option B”.

Note: To do this question, you need to know the permutations and combinations topic properly. Without using the formulas, you can also verify the options by taking some values for m and n. Then you can find the answers, by substituting them in the options.