AMC12 2008 12B, Problem 22

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?

Solution: This solution is pretty similar to the above. So there are 4 spots remaining, and it would be hard to count all the combinations where Auntie Em could park, but you can count all the combinations where Auntie Em can't park. Since the 12 cars are indistinguishable, we can use distributions here. There must be $a_1 \geq 0$ cars, then an empty spot, then $d_2 \geq 1$ cars, then an empty spot, then $d_3 \geq 1$ cars, then an empty spot, then $d_4 \geq 1$ cars, then an empty spot, then $a_5 \geq 0$ cars. \[a_1 + d_2 + d_3 + d_4 + a_5 = 12\] To remove all the restrictions, let $a_n = d_n - 1$. \[a_1 + a_2 + a_3 + a_4 + a_5 = 9\] We can now use stars and bars on these values (9 "stars", and 5 - 1 = 4 dividers), to get $\binom{13}{4}$ possibilities where she can't park. There are $\binom{16}{4}$ possibilities in total. \[\frac{\binom{13}{4}}{\binom{16}{4}} = \frac{11}{28}\] Subtracting that from 1 to get the probability she can park, the correct answer is $\fbox{\; \frac {17}{28}}$.

Answer: \; \frac {17}{28}