Difficulty in understanding the solution to N = min{n>= 2: Un > Un-1}

Question

It is Example 3.28 on page 124 in Ross's book (Introduction to Probability Models-11th edition)

Let $U_1$,$U_2$,... be a sequence of independent uniform $(0,1)$ random variables, and let $N$ = min{n >= 2: $U_n$ > $U_{n-1}$}

That is, N is the index of the first uniform random variables that is larger than its immediate predecessor.

The Solution Says:

It is easy to find the distribution of N. Since all $n!$ possible orderings of $U_1$,...,$U_n$ are equally likely, we have

P(N > n) = P{$U_1$ > $U_2$ > ... > $U_n$} = $1/n!$

My Questions:

1. Is "all $n!$ possible orderings" the sample space? Is the sequence [$U_1 = U_2 = ... = U_n$] in the sample space?

   I think the sample space contains cases where at least 2 of them 
   have the same value. I.E. in case of 2 Bernoulli variables the
   sample space is ([1,1],[1,0],[0,1],[0,0])
   

2. Why P{$U_1$ > $U_2$ > ... > $U_n$} = $1/n!$?

   Is n in P(N > n) countable or not countable? Does that really matters?
   My example: [1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]
   P(N > 2) = 2/6 ([2,1,3],[3,1,2])
   [3,2,1] this one is very strange
   

Your help will be great appreciated.

Answer 1

Assuming the $U_i$'s are continuous uniform variables, the case where any of them are equal (which would mess up the nice combinatorics, you're right) has probability $0$, and can thus be excluded from any calculation of probability. At least until you start conditioning on events with probability $0$, which we aren't doing.

As for why $P\{U_1 > U_2 > \ldots > U_n\} = 1/n!$, consider rewriting the situation from

Pick $U_1$, then pick $U_2$, then ..., then pick $U_n$

to

Pick $n$ random numbers. Then assign each of those numbers, randomly, to each of the random variables $U_1, \ldots,U_n$

We see clearly in the last case that the probability that $U_1$ gets assigned the largest number, $U_2$ the second largest, and so on, is $1/n!$.