Is my thought process correct for these permutations/combinations? I'm not very strong with math so be gentle. I decided against posting this to StackOverflow, even though it has to do heavily with Excel VBA. I think my primary questions belong here.
Let's say I have the numbers 1 through 30 (inclusive), and I'm "matching" them with each other and want to find out a few things:


*

*I'm confident with this, but want confirmation: am I correct in assuming that there are 30! possible ways to arrange the numbers 1 to 30? This feels like the deck of cards situation to me.


The next question is more difficult to describe so I'll attempt to simplify it.
Let's say instead that I have 30 people. I'm pairing people off every day: #1 pairs with #30, #2 pairs with #29, #3 pairs with #28, and so on...
Tomorrow they do the same thing, but since 1 and 30 already met, they can't anymore, ever again. #1 meets with #29 instead, and so on all the way down to #2, once per day until the options are exhausted.
In the end, I have decided that 29 possible outcomes exist, or n - 1. This is because each day you're removing an option that exists for every single person. So by the end of the 29 days, everyone will have met everyone else ONLY once.
My intention here is to produce all the possible outcomes in columns in excel, for n = 30 in this case, but ultimately for any n. I'm attempting to figure out the logic behind that, but wanted to confirm all the math first.
Thanks very much.
Edit: I have some follow up thoughts. The number 435 has been bugging me. I would argue that while 1 meets with 2, there are a number of other possible meetings available for the rest of the people. For example, 3 could meet with: 4, 5, 6...or 30. Does 435 cover all possible potential meetings? It may help to show the second part of my question here which shows a screenshot of partial results of my data: https://stackoverflow.com/questions/43684136/excel-vba-producing-all-combinations-of-a-range-of-values
Thanks again!
 A: You're right about the first part.
For the second part, 29 is indeed the correct answer (though your reasoning in itself only proves that it cannot be more than 29).

Here is a concrete way to construct a 29-day meeting schedule:
Appoint one of the persons king and give each of the rest of the persons a number from $1$ up to $29$.


*

*To find the day two non-royal will meet, add their numbers and subtract $29$ if the result is $30$ or more. The result is a number between $1$ and $29$, which you interpret as the number of a day.

*To find the day someone meets the king, add his number to itself and subtract $29$ if the result is $30$ or more.
(This works because every day each person can find someone to meet with that matches this rule: He subtracts his own number from the date and adds $29$ if that gives $0$ or a negative number. The result tells him whom to meet; if he gets his own number back, he will meet with the king.
Being the king is more complex: To find his partner he needs to divide either the date or the date plus $29$, by $2$ -- exactly one of those will be even).
This procedure works for every even number of persons. If we have an odd number of people, at least one of them need to be without a partner each day. The best we can do then is to pretend there is one person more (making an even number), and then cancel all of the pretend person's appointments after we make the plan. In that way it will take as many days as there are persons until everyone has met everyone, and everybody gets one day of rest along the way.

If you're going to implement this on a computer, the "subtract 29 if too large" business can be conveniently implemented by taking (x+y) mod 29 -- or (x+y) % 29, depending on language -- combined with numbering people and days from $0$ to $28$ instead of $1$ to $29$. (Since what is really going on behind the scenes here is artihmetic modulo $29$, calling number 29 number 0 instead will make no substantial difference).
