$\sum_{t=0}^n {2n \choose t + n} {t + n \choose n}=2^n{2n \choose n}$ I'm having trouble proving this combinatorially. I can't work the $2^n$ into my understanding. 
$$\sum_{t=0}^n {2n \choose t + n} {t + n \choose n} = 2^n{2n \choose n}$$
I'm thinking of it as taking $2n$ objects, splitting them into groups of size $t + n$ where $t$ are of one type and $n$ of another, and then splitting these groups into groups of size $n$ which mixes both types together and doing this for $t\leq n$. I can understand taking $2n$ groups and splitting them into groups of $n$ on the RHS as finding one such result but I don't see why it is done $2^n$ times.
 A: The left hand side counts the ways to pick $n$ or more out of $2n$ objects and then pick $n$ out of these. You might colour those final $n$ ones red, those $0\le t\le n$ ones picked in the first but not the second step blue.
The right hand side counts the ways to pick $n$ out of $2n$ objects to be painted red, and then pic an arbitrary subset of the remaining $n$ objects to be painted blue.
A: I think the algebraic proof is pretty straightforward.
$$\sum_{t=0}^n {2n \choose t + n} {t + n \choose n} =$$
$$={2n \choose   n}\sum_{t=0}^n  {n  \choose t} = 2^n{2n \choose n}.$$
If you want a combinatorical reasoning, consider $2n$ objects. You choose $t+n$ of them. Then, from these $t+n$ objects you choose $n$. Clearly, you obtain the same $n$-uplet several times (there're $\binom{2n}{n}$ different $n$-uplets, as you've already said). 
Let's count how many $t+n$-uplets can contain a fixed $n$-uplet. Apparently, the same number as we can have different $t$-uplets from a set of yet non-used objects (there're $n$ of them). It's easy to see that this is the same as counting the number of different subsets of a set with $n$ elements, which gives us the factor $2^n$.
A: Consider $2n$ distinct objects and all the possible $n$ element sets from this. You have $\dbinom{2n}n$ of these. For each of this $n$ element set, write down all its subsets. The total number of subsets you would have written down is $2^n \dbinom{2n}n$.
Now consider a subset of size $n-t$, where $t \in \{0,1,\ldots,n\}$. Let us count how many times this would have been written down. The number of times one subset of size $n-t$ occurs as a subset of an $n$ element set is $$\dbinom{n+t}{t}$$
This is because to form an $n$ element set, we need to fill the remaining $n-(n-t)=t$ elements from the remaining $2n-(n-t)$ elements. Hence, the number of times a subset of size $n-t$ would have occurred is $\dbinom{n+t}{t}$.
Now the number of possible subsets of size $n-t$ from $2n$ elements is $\dbinom{2n}{n-t}$.  Hence, the total number of subsets written down is
$$\sum_{t=0}^n \dbinom{2n}{n-t} \dbinom{n+t}t = \sum_{t=0}^n \dbinom{2n}{n+t} \dbinom{n+t}n$$
Hence, we get that
$$2^n \dbinom{2n}n = \sum_{t=0}^n \dbinom{2n}{n+t} \dbinom{n+t}n$$
