Combinatorial Methods for seating every other row Problem 1:
If $n$ people are seated in a random manner in a row containing $2n$ seats, what is the probability that no two people will occupy adjacent seats?
I know that the probability that $n$ people are seated in $2n$ rows is ${2n\choose n}$
I also know that the answer to this problem is $n+1\over {2n\choose n}$
Why does $n+1$ solidify that the people will not be sitting next to each other? I somewhat understand it, but I would appreciate a more logical explanation.
Problem 2:
What is the probability that a group of $50$ senators selected at random will contain one senator from each state?
So obviously the set of $50$ senators is the combinatorial ${100\choose 50}$ and we want $1$ senator from each state so that's ${2\choose 1}$.
So we have ${2\choose 1}\over {100\choose 50}$, but the ${2\choose 1}$ should be ${2\choose 1}^{50}$ for all 50 states. Why do we have to put the numerator to the 50th power? Shouldn't ${2\choose 1}$ suffice that each senator should be from $1$ state?
Thanks
 A: ad 1: $2n\choose n$ is the number of ways to seat $n$ people without regarding their identities. Now, how many of these will have no two people adjacent? You could occupy all odd seats, leaving the even ones free. Note that this will also leave seat $2n$ free, although this is not necessary in order to "keep the distance". So, you have one additional unoccupied seat which you can move about. It can be to the left of any person ($n$ possibilities), or to the right of the rightmost person ($1$ possibility). Therefore, you have $n+1$ ways to seat $n$ people.
ad 2: You can select senators state by state. For the first state, there are $2\choose 1$ possibilities to select exactly one senator. Likewise for the second state, and so on.
A: For problem 2, why not consider the smaller problem where there are only two states? Small enough that you can write down all the possibilities and see what the answer is, and then compare to the formulas you are asking about. 
A: Let us consider $6$ people and $3$ seats. Here is how we can arrange such that no two people are adjacent:
Let 3 people sit on any of the $2n$ seats:
$\#P_{1}\#P_{2}\#P_{3}\#$
Now we have to fill the remaining $2n-n=3$ chairs.
Note that there are $n-1=2$ spaces between three people.
Moreover, there are $n+1=4$ # seats between the three people. Therefore, 3 people can be arranged in $4$ ways
Lastly, each of the three people can be arranged in $3!$ ways.
Possible number of arrangement= $\frac{4 \times 3!}{6C3}$
Generalize the idea:
$\frac{(n+1) \times n!}{2nCn}$
