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How many ways to select $N$ things out of $2N$ things where $N$ things are alike and other $N$ things are unlike ?

I am suspecting something like :-

First choose $K$ things from $N$ unlike things in $C(N,K)$ and now its turn for $N-K$ things to be choosen from $N$ alike things.

Answer is given $2^N$ but I have no idea how to get that from my approach .

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    I don't get the question. If $N$ things are alike and other $N$ are alike, then we can choose $N$ as $N+0, (N-1)+1, ... , 0+N$ where $A+B$ means we took $A$ of the first kind and $B$ of the second. Therefore, $N+1$ way.Ah, unlike. Then it is obvious - to choose $N$ things is the same as choose any subset of $N$ unlike things and then add an appropriate number of alike. But there are $2^N$ subsets of $N$-element set. – 2017-01-29
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    Does 'likeness' here even matter? The question doesn't ask for you to choose the 'alike' or 'unlike' objects only, so each object can either be chosen or not. Also, once you have chosen $N$ objects, you can stop. Hence there are two choices for each of the $N$ objects and so the answer is $$\underbrace{2\times2\times2 \dots}_{\text{N times}} = 2^{N}$$ – 2017-01-29

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A selection which contains $N$ distinct things.

$\dbinom{N}{N}$

A selection which contains $1$ element of alike things and $(N-1)$ elements of distinct things.

$\dbinom{N}{N-1}$

A selection which contains $2$ element of alike things and $(N-2)$ elements of distinct things.

$\dbinom{N}{N-2}$

$\cdots$

A selection which contains $N$ element of alike things

$1$

So, required count

$\dbinom{N}{N}+\dbinom{N}{N-1}+\dbinom{N}{N-2}+\cdots+1 = 2^n$

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    Thanks !! This is so better :-) – 2017-01-29
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Yes, your approach is exactly right. It is first choosing $k$ unlike things, which can be done in $\binom{N}{k}$ ways and then choosing remaining $n-k$ like things, which can be done in only one way, thus one choice overall can be done in $\binom{N}{k}$ ways. Now, we sum the choices from $1$ to $N$ , that is, the answer is $\sum_{k=1}^N\binom{N}{k}\cdot1=2^N$

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    But, I am not getting why to sum the choices . – 2017-01-29
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    @JonGarrick It is the sum of binomial coefficients. – 2017-01-29
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There are N things to be chosen, each one can be of either type. Therefore for first one we have two choices,for second one again two choices, and so on.Therefore by multiplication theorem number of ways of doing so are 2*2*2...N times giving 2^N as answer.