Suppose I have the set of numbers $[{1,2,3...,10}]$
I (think) I know that if I choose a number randomly, the probability of choosing the $10$ different numbers of the set in $10$ tries is $\frac{10!}{10^{10}}$. This is because on the first try I dont care what number comes out, so the probability of not repeating is $\frac{10}{10}$, but on the second try I dont want to get the first number that I got, so the probability of not repeating is $\frac{9}{10}$ and so on.
My question is, what is the probability of randomly choosing all the numbers of a set that contains $n$ numbers over $X$ tries?
This probability can be computed using the principle of inclusion exclusion to be $$ \sum_{k=0}^n(-1)^k\binom{n}k\left(\frac{10-k}{10}\right)^X $$ It helps to understand this by looking at examples for small $n$.
When $n=1$, the probability is $1-(9/10)^X$. That is, the probability that you do get the one element you want is one minus the probability that you miss it $X$ times in a row.
When $n=2$, it is $1-2(9/10)^X+(8/10)^X$. There are two numbers $a$ and $b$ in our set. The probability we get both of them seems like it could be calculated by taking $1$, and subtracting the probability that $a$ is not obtained, $(9/10)^X$, and subtracting the probbaility $b$ is not obtained, $(9/10)^X$. However, outomces where both $a$ and $b$ are missed have been doubly subtracted, so these must be added back in. This accounts for the $(8/10)^X$ term.