Choosing $n$ objects in $2^{2n}$ ways Of $3n+1$ objects, $n$ are indistinguishable, and the remaining ones are distinct.  How that one can choose from them $n$ objects in $2^{2n}$ ways.
 A: The task is the same as selecting at most $n$ objects from the $2n+1$ distinguishable objects as these can be filled up with some of the indistinguishable objects to get $n$ objects.
Given a subset of a set of $2n+1$, exactly one of this set and its complement has size $\le n$. Therefore the desired number is precisely half the number of subsets of a $(2n+1)$ element set, i.e. $\frac12\cdot 2^{2n+1}=2^{2n}$.
A: Suppose that you choose $k$ of the distinguishable objects; you can do this in $\binom{2n+1}k$ ways. Once you've done that, can choose $n-k$ of the indistinguishable objects to fill out your set of $n$ objects in only one distinguishable way, since those objects are indistinguishable. Thus, there are $\binom{2n+1}k$ ways to choose such a set. The total number of ways to choose a set of $n$ objects is just the sum over $k$:
$$\sum_{k=0}^n\binom{2n+1}k\;.$$
This is exactly half of the subsets of a set of $2n+1$ elements; why?
A: Hint:
$\text{We can choose $n$ objects in
 }{2n+1\choose n}+{2n+1\choose n-1}+\cdots+{2n+1\choose 1}+{2n+1\choose 0}\text{ ways.} $
