What is the probability of randomly selecting n random numbers in the range 1-m in sorted order? For example say m is 10 and n is 3. I am selecting 3 numbers from 1-10 and want to know the probability with which they will be in sorted order. 1, 1, 4 would be fine. 1, 6, 3 would not be fine. For m=10 and n=2, the probability of selecting the 2 numbers in sorted order is 55/100. 
How could I calculate the probability for any arbitrary n and m?
 A: Recursive Solution
Let the first number chosen be $k$ (with probability $1/m$).  The chance that all are in order now equals the chance that the remaining $n-1$ are (a) in order and (b) equal or exceed $k$.  Subtracting $k-1$ from all of the remaining numbers puts them in one-to-one correspondence with the possible ways of selecting numbers in order from $1, 2, \ldots, m-k+1$; according to part (b), this chance has to be multiplied by $(m-k+1)^{n-1}$.  Letting $p(n,m)$ denote the chance, this provides the recursion
$$p(n,m) = \frac{1}{m}\sum _{k=1}^m \left(\frac{m-k+1}{m}\right)^{n-1} p(n-1,m-k+1),$$
with $p(1,m)=1$ to get it started.
The unique solution is
$$p(n,m) = \frac{(n+1)^{[m-1]}}{m^n(m-1)!}$$
where "$^{[m]}$" denotes an ascending factorial power; in general,
$$x^{[m]} = x(x+1) \cdots (x+m-1).$$
To verify the solution we need only to show it satisfies the recursion and the initial condition; this is a matter of algebraic checking.
For instance, with $n=3, m=10$ we obtain
$$p(3,10) =\frac{(3+1)^{[10-1]}}{10^3(10-1)!} = \frac{4\cdot 5\cdots 11 \cdot 12 }{10^3 (9 \cdot 8 \cdots 3 \cdot 2 \cdot 1)} = \frac{10 \cdot11\cdot 12}{10^3(3\cdot2\cdot1)}=\frac{11}{50}=0.22.$$

Combinatorial Solution
A selection of $n$ values (with repetition) from the numbers $\{1,2,\ldots, m\}$ is a sequence $(k_1, k_2, \ldots, k_n)$.  Such sequences are in one-to-one correspondence with the sequences $(k_1, k_2+1, \ldots, k_n + n-1)$ drawn from the numbers $\{1, 2, \ldots, m+n-1\}$.  (For instance, drawing $(1,1,4)$ from $1..10$ would correspond to drawing $(1,2,6)$ from $1..12$.)  Moreover, the original sequence is in order if and only if the derived sequence is in strict order.  It thereby determines (and is determined by) the subset $\{k_1, k_2+1, \ldots, k_n+n-1\}$, of which there are $\binom{n+m-1}{n}$ possibilities (by definition).  Because there are $m^n$ equally probable sequences, the desired probability is
$$p(n,m) = m^{-n}\binom{n+m-1}{n}.$$
This of course is just another way to express the previous formula for $p(n,m)$ (or, if you like, equating the two results gives us an explicit formula for the binomial coefficient!).
For example,
$$p(3, 10) = 10^{-3}\binom{3+10-1}{3} = 10^{-3}\binom{12}{3} = \frac{12\cdot 11\cdot 10}{10^3(3 \cdot 2 \cdot 1)} = \frac{11}{50},$$
exactly as determined with the recursive solution.
A: Unless I got something wrong, I think it might me possible to bring down you $n,m$ problem to a $2,n$ problem, which you have resolved for $n=10$ and of which the solution is more generally $p = \frac{n+1}{2n}$
$P[X_m \ge X_{m-1} \ge ... \ge X_1]=$
$P[X_m \ge \max (X_{m-1},X_{m-2}  ...  X_1);X_{m-1} \ge \max (X_{m-2},X_{m-3}  ...  X_1);...X_1 \ge \max(X_1)=X_1]$ 
Because they are $i.i.d$ this makes it:
$=P[X_m \ge \max (X_{m-1},X_{m-2}  ...  X_1]P[X_{m-1} \ge \max (X_{m-2},X_{m-3}  ...  X_1]... P[X_1 \ge \max(X_1)=X_1]$ 
$=\prod_{i=1}^{i=m-1} P[X_m \ge X_{i}] \prod_{i=1}^{i=m-2} P[X_{m-1} \ge X_{i}]\prod_{i=1}^{i=m-(m-1)} P[X_{2} \ge X_{i}]$
where $P_{i\not=j}[X_i \ge X_j] = \frac{n+1}{2n}$
