Confusion regarding probability problem involving 3 archers In a recent post , the following problem was asked. There was a solution provided; however, I was unable to fully comprehend it. I have copied verbatim the part that I was confused about here -
Problem
Alice, Bob and Carol are playing a game. Each round they press a number, which generates a random number independently uniformly distributed between $[0,1]$. If the number generated by this player is smaller than the smallest number so far, the player survives. Otherwise they get eliminated. They play in turns, i.e. Alice goes first, then Bob, then Carol, then Alice again etc.
What's the probability that Alice eventually wins this game?
Partial Solution
Let $a$ be the number of surviving rounds before the first eliminating round, and $b$ be the number of surviving rounds between the two eliminating rounds. Thus the game ends in $a + b + 2$ rounds. We consider the probability of this situation happening.
If we generate $a + b + 2$ random numbers, then their order take equal probability $\frac 1{(a + b + 2)!}$ for each possible permutation. Among all $(a + b + 2)!$ such permutations, only $a(a + b + 1)$ of them correspond to the situation above, namely the $(a + 1)$-th number can be inserted anywhere before the smallest of the first $a$ numbers, and the $(a + b + 2)$-th number can be inserted anywhere before the smallest of the first $(a + b + 1)$ numbers.
This means that the probability that one player gets eliminated on the $(a + 1)$-th round and another player gets eliminated on the $(a + b + 2)$-th round is equal to $\frac{a(a + b + 1)}{(a + b + 2)!}$.
My Question
For the above posted answer, why is it the case  that the $a(a+b+1)$ cases correspond to the given situation? For example, why can we insert the $(a+1)$-th number anywhere before the smallest of the first $a$ numbers? What if the first $a$ numbers were not ordered in a decreasing sequence? That would mean that there is an elimination round before the $(a+1)-$th round, which is a contradiction. So we cannot simply insert the $(a+1)-$th number as suggested above without considering more restrictions, right? Or am I missing something obvious here?
 A: After you generate $a+b+2$ random numbers, we can assign each of them a rank. A number's rank tells us the position of that number when all of them are arranged in ascending order.
For a valid permutation (i.e. we get $a$ rounds before the first elimination and $b$ rounds between the two eliminations), you need the first $a$ numbers to have strictly decreasing ranks. But the $a + 1$-th number can have any rank larger than the smallest of the first $a$ numbers, which allows for $a$ possible ranks ("insertions in the ascending queue"). Similarly, the next $b$ numbers have to be in a strictly decreasing rank order, lower than the lowest rank of the first $a + 1$ numbers too. Hence, for the $a + b + 2$-th number, it can have any rank larger than the smallest number so far, i.e. the smallest number of the $a +b+1$ numbers so far. It allows the $a+b+1$ possible ranks ("insertions") for this number in the queue.
ADDENDUM
For example, let $a = 2$ and $b = 1$. Let us call the numbers we sampled $x_1, x_2, x_3, x_4, x_5$. Now, for a valid permutation, we need the following:
\begin{align}
x_1 > x_2 && \text{no elimination at round 2}\\
x_3 > x_2 && \text{elimination at round 3} \\
x_4 < x_2 && \text{no elimination at round 4} \\
x_5 > x_4 && \text{elimination at round 5} \\
\end{align}
Of the overall descending sequence, we have $x_1 > x_2 > x_4$ fixed. Since $x_3$ is greater than $x_2$ and $x_4$, it has two positions to be in this sequence, either immediately before $x_2$ or before $x_1$. But $x_5$ needs to only be greater than $x_4$, which gives it 4 positions for each placement of $x_3$, i.e. it can be immediately before either of the $x_1, x_2, x_3$ or $x_4$.
