Probability that no car is parked next to a car of the same type A university's faculties have 30 cars of exactly 3 types: BMW, Toyota and Mazda.  University has build a new parking lot which has 30 parking slots. Each faculty  member randomly arrives to the parking lot till 11:00 AM and parks his car in any available slot. What is the probability that at 11:00 AM each parking slot has different type of car next to it?(That is no BMW is next to a BMW and no Toyota is next to a Toyota and no Mazda is next to a Mazda)
My attempt to solve this:
number of elements in Sample Space
  = number of arrangements of 30 cars in 30 slots
  = 30(Permutation)30
  = 265252859812191058636308480000000

I just was not able to count my events i.e., each parking slot has different type of car next to it.
What should I do?
 A: My humble try:
Let us create 30 spots 1 2 3 4 ....   30.  Assumption is each spot could be filled next by any of the three car types of equal probability.
The probability that any of the three cars could be chosen for the 1st spot is $=\frac{1}{3}$
Then the probability that the second spot is filled with car that is not the previous one = $=\frac{2}{3}$.  Having filled the first two spots.  The probability that the remainder of the spots will be filled with one that is not the same as the previous one is all equal to $=\frac{2}{3}$
Thus it is $=\frac{1}{3}\frac{2}{3}\cdots \frac{2}{3}$ that is one $\frac{1}{3}$
 and 29 $\frac{2}{3}$'s
.  This 1/3 could be placed in 10 different ways.
Hence the probability that no two adjacent cars are alike is =
$$\frac{2^{29}}{3^{30}}.10 =   0.000026075 $$
A: We first count the number $N$ of admissible strings containing $10$ each of $B$, $M$, $T$. One third of these strings begins with a $B$. Such strings have one of the two forms
$${\rm (a)}\quad Bw_1Bw_2\ldots Bw_{10},\qquad{\rm(b)}\quad Bw_1Bw_2\ldots Bw_{9}B\ ,$$ whereby the subwords $w_i$ are nonempty and consist of alternating  $M$'s and $T$'s.
(a) Assume  that $r$ subwords $w_i$  contain at least one letter $M$ (these $w_i$ can be chosen in ${10\choose r}$ ways), while the remaining $w_i$ consist of a single $T$.   There are $10-r$ letters $M$ left to distribute among the $r$ chosen $w_i$, and this can be done in ${10-r+r-1\choose r-1}={9\choose r-1}$ ways. Each of these thrown in $M$'s has to be preceded by a $T$ in order to separate it from the foregoing $M$. Furthermore we need  $10-r$ more letters $T$ for the remaining $w_i$. In all we now have used all the $B$'s and $M$'s, and $(10-r)+(10-r )=20-2r$ of the $T$'s, so that  $2r-10$ letters $T$ remain to be distributed (this of course enforces $r\geq5$). These $T$'s can be placed at the head or the tail of a $w_i$ already containing $M$'s, hence there are $2r$ slots for these $T$. It follows that we can place the remaining $T$'s in ${2r\choose2r-10}$
ways. The total number $N_a$ of strings of type (a) is therefore  given by
$$N_a=\sum_{r=5}^{10}{10\choose r}{9\choose r-1}{2r\choose 2r-10}=28\,964\,128\ .$$
(b) Assume  that $r$ subwords $w_i$  contain at least one letter $M$ (these $w_i$ can be chosen in ${9\choose r}$ ways), while the remaining $w_i$ consist of a single $T$.   There are $10-r$ letters $M$ left to distribute among the $r$ chosen $w_i$, and this can be done in ${10-r+r-1\choose r-1}={9\choose r-1}$ ways. Each of these thrown in $M$'s has to be preceded by a $T$ in order to separate it from the foregoing $M$. Furthermore we need  $9-r$ more letters $T$ for the remaining $w_i$. In all we now have used all the $B$'s and $M$'s, and $(10-r)+(9-r )=19-2r$ of the $T$'s, so that  $2r-9$ letters $T$ remain to be distributed (this of course enforces $r\geq5$). These $T$'s can be placed at the head or the tail of a $w_i$ already containing $M$'s, hence there are $2r$ slots for these $T$'s. It follows that we can place the remaining $T$'s in ${2r\choose 2r-9}$
ways. The total number $N_b$ of strings of type (b) is therefore  given by
$$N_b=\sum_{r=5}^{9}{9\choose r}{9\choose r-1}{2r\choose 2r-9}=12\,685\,428\ .$$
In this way we obtain
$$N=3(N_a+N_b)=124\,948\,668\ .$$
The probability $p$ in question therefore computes to
$$p={N (10!)^3\over 30!}={800\,953\over35\,583\,312\,765}\doteq0.0000225092
\ .$$
Counting 250 workdays per year this means that we could expect one such event every 177.7 years.
