smc

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 $a_1\ge 0$ cars, then an empty spot, then $d_2\ge 1$ cars, then an empty spot, then $d_3\ge 1$ cars, then an empty spot, then $d_4\ge 1$ cars, then an empty spot, then $a_5\ge 0$ 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 $\left(\genfrac{}{}{0ex}{}{13}{4}\right)$ possibilities where she can't park. There are $\left(\genfrac{}{}{0ex}{}{16}{4}\right)$ possibilities in total. $$\frac{\left(\genfrac{}{}{0ex}{}{13}{4}\right)}{\left(\genfrac{}{}{0ex}{}{16}{4}\right)}=\frac{11}{28}$$ Subtracting that from 1 to get the probability she can park, the correct answer is $\boxed{\frac{17}{28}}$.