Among $2n$ items, $n$ are the same. How many ways are there to choose $n$ items from $2n$ items? Among $2n$ items, $n$ are the same. How many ways are there to choose $n$ items from $2n$ items?
So, first thought is: n are the same and n are different.
I can't understand how we get to $2^n$ with this solution:
The solution says:


*

*We choose with $1$ way $n$ of the $n$ same items and with ${{n}\choose{0}}$ ways $0$ of the $n$ different items.

*We choose with $1$ way $n-1$ of the $n$ same items and with ${{n}\choose{1}}$ ways $1$ of the $n$ different items.

*We choose with $1$ way $n-2$ of the $n$ same items and with ${{n}\choose{2}}$ ways $2$ of the $n$ different items.

*...

*We choose with $1$ way $0$ of the $n$ same items and with ${{n}\choose{n}}$ ways $n$ of the $n$ different items.


And end with summing:
$$ {{n}\choose{0}} +{{n}\choose{1}} +{{n}\choose{2}} + \dots + {{n}\choose{n}} = \sum\limits_{i=0}^n {{n}\choose{i}} = 2^n  \text{ ways}$$
 A: Your idea can be carried out without mentioning binomial coefficients. We have $2$ boxes of balls. Box 1 has $n$ distinct balls, and Box 2 has $n$ identical balls.
Call the set of balls in Box 1 by the name $A$.  
To choose exactly $n$ balls, we choose any subset of $A$. If we have not chosen all the balls $A$, we grab enough balls from Box 2 to make up the total of $n$.
So there are exactly as many ways to carry out our task as there are ways to choose balls from Box 1. But the set $A$ has $2^n$ subsets, so there are $2^n$ ways to carry out our task.
A: There are $2^n$ ways to pick some elements from the $n$ non-identical ones (there are $2^n$ subsets of the $n$ non-identical elements set). Then you can complete these elements to $n$ elements, by adding some identical elements.
It is easy to justify that this construction produces all sets and each of them uniquely.
A: The solution is true, maybe you don't understand the logic behind it. 
$-$ First we choose $n$ of the same $n$ items, which leaves us with choosing 0 from the other $n$ "not-same" items. Beacuse the order is not important it can be caluclate using the combination formula $\binom{n}{k}$, in this case $k = 0$.
$-$ Second we choose $n-1$ items of the same $n$ items, which means we have to chose 1 from the other $n$ items. We calculate all these combination with the same formula, but this time $k = 1$.
$$...$$
$-$ And at last we chose 0 of the same $n$ items, which means we have to choose $n$ items from the other half. And all the combinations are given by the same formula.
Finally we add all binomial coefficient together and we get the identity:
$$ {{n}\choose{0}} +{{n}\choose{1}} +{{n}\choose{2}} + \dots + {{n}\choose{n}} = \sum\limits_{i=0}^n {{n}\choose{i}} = 2^n$$
The sum of all binomial coefficient is the same of all numbers in the $n^{th}$ row in the Pascal triangle. And we now that the sum of of all numbers in the $n^{th}$ row in the Pascal triangle is given by $2^n$.
I assume that you know how we obtain the Pascal triangle. If the sum in one row is $n$ the the sum of the next row is $2n$, that's because every number of the first row impacts two numbers of the next row. We know that the sum of the first row (we are counting from 0, so n = 0) in the Pascal's triangle is 1, the second row sum is 2, the third 4... And for the $n^{th}$ row the sum is $2^n$
You could even think like this. We'll observe only the n "non-same" numbers. There are two possibilities, it's either included in the $n$ chosen items or not. So we have 2 option for every item. Because the events are independent from each other the number of ways is:
$$2 \times 2 \times ... \times 2 = 2^n$$
If we choose $k$ items of the "non-same" items, then we just simply chose $n-k$ same items in order to choose $n$ items in total.
