Ordering students in a row so that Danny is to the right from a certain group of students

Question

So if I have a row in which I need to order n students so that Danny (one of the n students) sits to the right from Danna (another student), the answer will be: $\frac{n!}{2}$

This makes intuitive sense because we can understand that out of all possible permutations, half of them, Danny will be to the right of Danna and the other half, Danny is on the left of Danna (another student).

This is very nice but what if we want to know how many permutations are there for Danny being to the right from Danna and Dinna?

This really confuses me because I can't treat Dinna and Danna as a pair like in many combinatorics exercises, because of possible permutations like this:

$$Dinna, ... ,John, Johnson, Danna, Danny$$

This breakes my head and I can't help not to wonder about what if I need Danny to sit to the right from an x amount of students. ($Let$ $x<n$)

Answer 1

Danny and the other named students will (in the end) occupy a certain set of chairs. Imagine all other (anonymous) students are seated first and then you choose someone to go in the rightmost chair of the remaining set. Then if there are $x$ students that you would prefer to see Danny to the right of, there is a $\dfrac{1}{x+1}$ chance that that will happen, by choosing Danny for that chair rather than one of the other $x$ named students.

So to find out the total number of arrangements in which Danny is to the right of $x$ other students, we simply divide the total number of arrangements by $x{+}1$, that is, $\dfrac{n!}{x+1}$. If there is only one other student of interest, this of course calculates out as $\dfrac{n!}{2}$ arrangements meeting the constraint.

Answer 2

There are $n!$ orders in total.

The number of orders with Danny at left (of Danna and Dinna) equals the number of orders with Danny in the midst and also equals the number of orders with Danny on the right.

So in $\frac{n!}{3}$ of the orders Danny will be at left.