Q:

Probability of choosing $n$ numbers from $\{1, \dots, 2n\}$ so that $n$ is 3rd in size

We uniformly randomly choose $n$ numbers out of $2n$ numbers from the group $\{1, \dots, 2n\}$ so that order matters and repetitions are allowed. What is the probability that $n$ is the $3^{\text{rd}}$ number in size in the chosen series? (= there are only two bigger than $n$)Note that if a number that is bigger than $n$ was chosen more than once, we still count it as one number bigger than $n$.
Example: $n = 5, \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ 
$7, 1, 5, 9, 9$ - ok
$2, 2, 10, 5, 7$ - ok
$6, 6, 3, 5, 5$ - not ok, 5 is $2^{\text{nd}}$ in size
My lead was to first choose 2 numbers that are bigger than $n$ out of the n numbers that are bigger than $n$ in the group $\{1, \dots, 2n\}$. Then in order to complete a series of $n$ numbers I need to choose $n-3$ more from the numbers that are equal or smaller than $n$, but then I figured this is not right as a series could be formed from the 2 numbers bigger than $n$ and $n$ itself without any additional numbers. At this point I have no additional leads and could very much use your assistance and guidance. 

A:

You can use the inclusion-exclusion principle.
Let $S:=$ The event of getting a valid sequence of numbers ($n$ numbers $\le n$ and 2 numbers $\gt n$)
$A_1 :=$ The event that $n$ does not appear in the sequence
$A_2 :=$ The event that one of the larger numbers than $n$ does not appear in the sequence
$A_3 :=$ The event that the second number that is larger than $n$ does not appear in the sequence
Now, the probability we are looking for is the probability that S occurs but the union of $A_1,\, A_2,\, A_3$ does not occur.
We have the formula
$$ \Pr \left[S\setminus \bigcup_{i=1}^{3}A_i \right] = \Pr \left[ S \right] - \sum_{i=1}^{3} \Pr\left [ A_i \right ] + \sum_{1\leq i< j\leq 3} \Pr\left [ A_i\cap A_j \right ] - \sum_{1\leq i<j<k\leq 3} \Pr\left [ A_i\cap A_j\cap A_k \right ] $$
Let's calculate the probabilities of the events defined above.
First of all, our sample space $\Omega$ is all sequences of $n$ numbers from the group $\{1, 2, \ldots, 2n\}$.
So $\left| \Omega \right| = \left( 2n \right)^n $.
Now, $ \Pr \left[S \right] = \frac{\left ( n+2 \right )^n}{\left ( 2n \right )^n} $ (we have $n$ numbers $\le n$ and 2 numbers greater than $n$)
Similarly, $ \Pr \left[A_1 \right] = \Pr \left[A_2 \right] = \Pr \left[A_3 \right] = \frac{\left ( n+1 \right )^n}{\left ( 2n \right )^n} $
We also have to calculate the intersections.
$ \Pr\left [ A_i\cap A_j \right ] = \frac{n^n}{\left ( 2n \right )^n} \; ($for $ 1\leq i<j\leq 3 )$ - notice there are 3 such intersections. And also there is the intersection of all three:
$ \Pr\left [ A_1\cap A_2\cap A_3 \right ] = \frac{\left ( n-1 \right )^n}{\left ( 2n \right )^n} $
One last thing - there are $ \binom{n}{2} $ ways to choose those 2 numbers that are greater than $n$.
Now, we'll plug everything in the inclusion-exclusion formula and multiply by $ \binom{n}{2} $, and we'll get:
$$ \Pr \left[S\setminus \bigcup_{i=1}^{3}A_i \right] = \binom{n}{2} \left( \frac{\left ( n+2 \right )^n}{\left ( 2n \right )^n} - 3\frac{\left ( n+1 \right )^n}{\left ( 2n \right )^n} +3\frac{n^n}{\left ( 2n \right )^n} - \frac{\left ( n-1 \right )^n}{\left ( 2n \right )^n} \right) = \binom{n}{2} \frac{\left ( n+2 \right )^n - 3\left ( n+1 \right )^n + 3n^n - \left ( n-1 \right )^n}{\left ( 2n \right )^n} $$

