What is the distribution of gaps? Randomly select $n$ numbers from the universe $\{1,2\dots,m\}$ with or without replacement, and sort the numbers in ascending order. 
We can get a list of number $\{a_1,a_2,\dots,a_n\}$, and then we can get the difference between two consecutive numbers and get the gap list:
$\{a_1, a_2-a_1,\dots ,a_n-a_{n-1}\}$
So my question is: what is the distribution of the gaps. 
Let $A_i$ be the number of gaps which are equal to $i$, what is the distribution of $A_i$? 
Update: I clarify that there should two situation to consider: with or without replacement.
 A: An inclusion-exclusion argument gives us the following for the distributions of the $A_i$.
We first consider the case of the $n$ numbers being distinct (selection without replacement).
If $1\leqslant n \leqslant m$, $1\leqslant i \leqslant m+1-n$ and $0\leqslant k \leqslant M_i$, then we have:
$$
\textrm{P}(A_i=k)\ =\ P\:(m,n,i,k)\ =\ \frac{1}{T}\: \sum_{j=k}^{M_i} (-1)^{j-k} {j\choose k} {n\choose j} {m-ij\choose n-j}
$$
where $T={m\choose n}$ is the total number of configurations, and $M_i=\textrm{min}(n,\lfloor\frac{m-n}{i-1}\rfloor)$ is the maximum possible gap size.
$Q_k={n\choose k} {m-ik\choose n-k}$ is a count of the combinations with at least $k$ gaps of size $i$, determined by first choosing $k$ gaps of size $i$ and then arbitrarily choosing how the remaining 'space' is split up. However $Q_k$ counts combinations with more than $k$ $i$-gaps multiple times; indeed combinations with $j$ $i$-gaps are counted $j \choose k$ times. Thus if $R_k$ is the number of combinations with exactly $k$ gaps of size $i$, we have $R_k=Q_k-\sum_{j=k+1}^{M_i}{j \choose k}R_j$. Inclusion-exclusion gives us that $R_k=T\times P\:(m,n,i,k)$.

If we now consider the case of the $n$ numbers not necessarily being distinct (selection with replacement), then, if we include $0$ in the set of numbers that may be selected (so that the first gap can be $0$ like the others), we have:
$$
\textrm{P}(A_i=k)\ =\ P\:(m+n,n,i+1,k)
$$
where $P\:$ is as above.
This follows simply from the fact that selecting $n$ numbers from $m+1$ with replacement is equivalent to selecting $n$ numbers from $m+n$ without replacement, but with each gap being smaller in size by $1$.
A: Here's a partial solution  when sampling without replacement:
In addition to the gaps $G_1=a_1$and $G_j:=a_j-a_{j-1}$ for $2\leq j\leq n$, 
it is convenient to introduce the final gap $G_{n+1}=(m+1)-a_n$.
Then the random vector $(G_1,G_2,\dots, G_{n+1})$ gives a random composition
of the number $m+1$. That is, all outcomes $(g_1,g_2,\dots, g_{n+1})$
 with $$g_1+g_2+\cdots+g_{n+1}=m+1,\quad g_j\geq 1$$
are equally likely. There are $m\choose n$ such compositions, as 
found using stars and bars. 
Let's focus on the number $N$ of gaps of size 1. How many compositions of
$m+1$ into $n+1$  parts have exactly $k$ ones? The answer is 
$${n+1\choose k}\,{m-(n+1)\choose n-k}.$$
The left hand binomial coefficient counts how many ways 
we can place the ones, and the right hand binomial coefficient counts
the number of compositions of $m+1-k$ into $n+1-k$ parts where 
all the values are $\geq 2$, via another stars and bars argument. 
Thus $$\mathbb{P}(N=k)={{n+1\choose k}\,{m-(n+1)\choose n-k}\over{m\choose n}}.$$
I haven't thought about larger gaps, but maybe this idea will 
be useful there as well. 
