We can generate random numbers using the following equation \begin{align} x_{n+1} = (ax_{n} + c) mod(m) \end{align} $x_{n}$ takes the values $1,2,\ldots, m-1$ and we take $x_{n}/m$ as the pseudo random number, which is uniformly distributed between $(0,1)$ for suitable choice of $a, c, m$.
Suppose we want to generate a permutation of integers from $1, 2, \ldots, n$ such that each of the permutations is equally likely. Assuming we have a uniform random generator $U$ with us, \begin{align} P(Int(kU) + 1 = i) &= P(Int(kU) = i-1) = P(i-1 \leq kU < i)\newline &= P(\frac{i-1}{k} \leq U < \frac{i}{k}) = \frac{1}{k} \end{align} which gives us randomly generated random integers between $1$ and $k$ with equal probability. An easy way to generate permutation is
Choose a permutation $r_{1}, r_{2}, \ldots, r_{n}$ which can just be $r_{j} = j$
Let $k = n$
Choose a random number $U$ and let $I = Int(kU) + 1$
Interchange numbers at position $k$ and $I$
$k = k-1$
if $k > 1$ goto step 3 else return permutation
The above algorithm can also be used to get a random subset of size $r$ from a set $1, \dots, k$ by simply running the algorithm till $k = r$ since the elements in the last $r$ positions can be selected. For $r > n/2$, we find the $k=n-r$ elements not in the subset.