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In how many ways can we select $n$ objects from a collection of size $2n$ that consists of $n$ distinct and $n$ identical objects?

The answer is $2^n$ and I really don't see how they get this. Selecting $n$ distinct from $2n$ is $\binom{2n}{n}$.

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Try working a small example: one white sock, one red sock, two black socks. Hint: sort the 12 possibilities in order by number of black socks. – Eric Lippert Dec 17 '13 at 0:31

Let $A=\{a_1,\ldots,a_n\}$, where the $a_k$ are mutually distinguishable, and let $S$ be $A$ together with $n$ indistinguishable objects. An $n$-element subset $X$ of $S$ is completely determined when you know $X\cap A$: if $|X\cap A|=k$, the remainder of $X$ is just $n-k$ of the indistinguishable objects. $A$ has $2^n$ subsets, so there are exactly $2^n$ different possibilities for $X\cap A$ and therefore for distinguishable sets $X\subseteq S$ of cardinality $n$.

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Hint. If you choose $k$ elements from the $n$ distinct, and $n-k$ from the $n$ identical, there are:

$$\binom{n}{k}\cdot 1$$

ways to do this. Now sum from $k=0$ to $k=n$.

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Note that the set of objects chosen is determined by which of the distinct objects are chosen and how many of the identical objects are chosen. For each of the $2^n$ subsets of the $n$ distinct objects, the number of identical objects we need to take in order to fill our quota of $n$ total objects is uniquely determined. Hence there are $2^n$ total ways to take $n$ total objects from $n$ distinct and $n$ identical objects.

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