The way to produce a random but ordered series of numbers?
Suppose you want $n$ random numbers $a_k$, uniformly distributed between $0$ and $1$.
The probability that $a_1\gt x$ is $(1-x)^n=c_1$, so pick a uniform random number $c_1$ and let $b_1=1-\sqrt[n]{c_1}$, and $a_1=b_1$.
Now you need $n-1$ numbers uniformly distributed between $a_1$ and $1$.
Let $b_2=1-\sqrt[n-1]{c_2}$ and $a_2=a_1+(1-a_1)b_2$.
In general, $$c_k=Rand() \\b_k=1-\sqrt[n+1-k]{c_k} \\
a_k = a_{k-1}+(1-a_{k-1})b_k$$
EDiT
Take just two numbers $x$ and $y$. All points $(x,y)$ in the unit square are equally likely. Once you know the lower number is $c$, you are restricted to the two lines $(c,y), c\le y \le1$ and $(x,c),c\le x\le1$. In either case, the points on the line segment are equally likely, so the remaining variable is uniformly distributed between $c$ and $1$.
With three numbers, you start with points $(x,y,z)$ in a cube. Once you know the lowest value, you are restricted to three squares at right angles. Within any one, say $(c,y,z)$, the remaining variables $(y,z)$ are uniformly distributed over a square, and just need rescaling to be uniformly distributed over the unit square.
It won't always be the case. There is a probability density function, or PDF, which plays the role of ordinary probability. Then calculus plays an important role.
You want to produce, for some given $n$, independent uniformly distributed random numbers $X_i\in[0,1]$ for $i=1,\ldots, n$, and then let $Y_1,\ldots,Y_n$ be the same numbers but in ascending order.
Note that $P(Y_1<c)=\prod_{i=1}^nP(X_i<c)=c^n$. As $P(\text{rand()}<c)=c$, you can produce $Y_1$ easily per $Y_1=\text{rand()}^{1/n}$. After that, note that the remaining numbers are uniformly distributed in $[Y_1,1]$. So by the same reasoning, we can let $Y_2=Y_1+(1-Y_1)\cdot\text{rand()}^{1/(n-1)}$ and so on, i.e., $Y_{k+1}=Y_k+(1-Y_k)\cdot\text{rand()}^{1/(n-k)}$.
Nevertheless, I suggest that in most circumstance the straightforward method of generating and sorting is preferable. After all, sorting takes only $O(n\ln n)$ time and we avoid the time for all those raising to the power (and possibly introduced inaccuracy).