combinatory

Question

A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

$\textbf{(A)} \; \frac {11}{20} \qquad \textbf{(B)} \; \frac {4}{7} \qquad \textbf{(C)} \; \frac {81}{140} \qquad \textbf{(D)} \; \frac{17}{28}$

Answer 1

12 cars will take 12 places and we are left with 4 places. So, we need to find the probability that at least 2 of them are adjacent. This will be equal to 1 - Probability that none of them are adjacent. 

Let the cars be represented by C and vacant space by V. 

C C C C C C C C C C C C V V V V taking 16 places.

We have 16 places to choose the 4 V's giving ¹⁶C₄ ways.

If we want no two V's to be adjacent, we must place the V's around the 12 C's in 13 places- 12 after each C and 1 before the first C. i.e.; we have ¹³C₄ ways where no vacant places are adjacent. 

So, our required probability = 1 - ¹³C₄/¹⁶C₄ = 1 - (13 * 12 * 11 * 10 / 16 * 15 * 14 * 13)
= 1 - 11 / 28
= 17 / 28