How many ways to select $N$ things out of $2N$ things? 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 .
 A: 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
$\dbinom{N}{N}+\dbinom{N}{N-1}+\dbinom{N}{N-2}+\cdots+1 = 2^n$
A: 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 $\sum_{k=1}^N\binom{N}{k}\cdot1=2^N$
A: 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.
