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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 : $$\frac{(n-k)!}{k!((n-k)-k)!}$$

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  • $\begingroup$ Welcome to Math.SE. Take a look at How to ask a good question at Math.SE. To avoid downvotes and closing you should add your own efforts to the question by means of an edit (not a comment), and tell us where you got stuck.Hint: How many ways are there to pick $k$ out of $n$ items? How many ways are there to pick $k$ items out of $n-k$ (unused) items? $\endgroup$
    – drhab
    Commented Oct 26, 2019 at 8:24

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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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  • $\begingroup$ The numerator is the number of ways that $k$ cards can be picked from the $n-k$ cards that where not picked the first time: $\binom{n-k}{k}=\frac{(n-k)!}{k!(n-2k)!}$. You write wrongly that there are $\frac{n!}{k!(n-2k)!}$ possibilities. E.g. let $n=4$ and $k=2$. Then there is only $1=\binom{4-2}{2}$ way to pick the $2$ cards that are not picked the first time. Not $\frac{4!}{2!0!}$ ways. $\endgroup$
    – drhab
    Commented Oct 27, 2019 at 15:53

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