Assuming that there is a $M \times N$ matrix whose entries are ${0,1}$ binary.
For each row, there are $K$ $1$s and $N-K$ $0$s, and the positions are randomly picked.
In this case, I would like to know the probability that the matrix has no pair of two same columns. I think the probability that there is at least 1 pair of columns which are exactly the same can also give the answer to this.
There are ${N \choose K}$ potential columns with $K$ ones and $N-K$ zeros.
There are therefore ${N \choose K}^M$ equally likely ways of choosing $M$ of them, of which $\frac{{N \choose K}!}{\left({N \choose K}-M\right)!}$ have all distinct columns, making the probability of all distinct columns $\frac{{N \choose K}}{{N \choose K}}\frac{{N \choose K}-1}{{N \choose K}}\frac{{N \choose K}-2}{{N \choose K}}\cdots\frac{{N \choose K}-(M-1)}{{N \choose K}}$ $$=\dfrac{{N \choose K}!}{{N \choose K}^M\left({N \choose K}-M\right)!}.$$
As a numerical example, if $N=14$ and $K=3$ and $M=23$ then this gives a probability of about $0.492$ (not far away from the birthday problem).