I have a box which contains $n$ unused items (all items in the box are unused). From them I randomly pick $k$, where $k < n$ items. Those $k$ items became used when I picked them, and then I put them back into the box. How to find the probability that second time all $k$ items I pick are going to be unused?(not those which I already picked before)
I already know that all possible ways of picking $k$ from $n$ is: $$\frac{n!}{k!(n-k)!}$$
and that possible ways of picking $k$ from $(n-k)$ is : $$\frac{(n-k)!}{k!((n-k)-k)!}$$
Answer
There are indeed $\binom{n}{k}$ ways of picking $k$ items out of $n$.
The number of possible ways of picking $k$ from $n-k$ is $\binom{n-k}{k}$ (so not $\frac{n!}{k!(n-2k)!}$).
These "ways" are equiprobable so the event of picking the second time $k$ items that are not picked the first times has probability:$$\frac{\binom{n-k}{k}}{\binom{n}{k}}$$
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