Probability that 2 of t strings of length n are equal Given t bit-strings of length n that are generated randomly.
What is the probability that at least 2 of these strings are equal.
I've seen someone who wrote that the probability is $ \le \frac{t^2}{2^n}$
The reasoning is that you have $t$ options to choose the first string.
The $t$ options to choose the second string, and given a string chosen at the first stage, the probability that the other chosen string is equal to the first one is $\frac{1}{2^n}$.
The you sum all these options of choosing 2 equal string ($t^2$) and you get $ \frac{t^2}{2^n}$
This just seems wrong because for each such choice of 2 strings there are many other options to the other strings, but I can't find a proof to show that it is wrong.
Does this reasoning make sense? 

I think this probability is $1-\frac{\binom{2^n}{t}}{\binom{2^n -1 +t}{2^n -1}}$. Essentially this means 1- probability to have all strings different. The numerator is all different strings. The denominator is all cases of selecting t out of $2^n$ with repetitions.
EDIT: 
OK @MishaLavrov's solution $\frac{\binom{2^n}{t} t!}{2^{nt}}$ is better. What my solution gives is $\frac{\text{number of ways to get $t$ unique strings length $n$ from a set of $2^n$ strings, order does not matter}}{\text{number of ways to get $t$ strings length $n$ from a set of $2^n$ strings, order does not matter}}$.
For example, if $t=3$, one possible denominator would be $aba, aab, baa$ counted as one. In the numerator we can have $abc,bac,...,cba$ all counted as one outcome. 
My confusion is that, even if strings are 'generated randomly', do we account for repeated sets or not? 
