Combinatorial proof for identity $\sum_{k=0}^{2n}(-1)^k\binom{3n}{k}=\binom{3n-1}{2n}$ Combinatorial proof for identity: $$\sum_{k=0}^{2n}(-1)^k\binom{3n}{k}=\binom{3n-1}{2n}$$
Attempt: I can see that the left side is the number of ways of picking a $2n$ subset of a set $S = \{1,2,\dots,3n-1\}$ but i have no idea how to aproach the right hand side with sign changing and $3n$ instead of $3n-1$. 
Edit: I think i have solved it: we define two sets $$P := \{A\subseteq [3n];|A|=2k; k \lt n;k\in\mathbb Z\}\cup\{A\subseteq [3n];|A|=2n,1\in A\}$$
and $$Q:=\{A\subseteq [3n];|A|=2k-1; k \le n;k\in \mathbb N\}$$
and a function:
$$f:P\to Q;\ f(A) := A\triangle\{1\}$$
We see that $f$ is 
$$f(A)=
\begin{cases}
A\cup\{1\},&|A|=2k,k<n,1 \not\in A\\
A\setminus\{1\},&|A|=2k,k\le n,1\in A\\
\end{cases}
$$
in other words $|f(A)|\in\{1,3\dots,2n-1\}$.
It can be checked that $f$ is a bijection whick pairs all the subsets on the LHS except the ones with size $2n$ that don't contain $1$. There are $\binom{3n-1}{2n}$ such sets.
 A: For the sake of clarity, maybe you could denote by $X_n$ the set $\{1,\ldots,n\}$ and write $\mathcal E_n:=\{S\subseteq X_n\mid |S|=2k, 0\le k\le n\}$ and $\mathcal O_n:=\{S\subseteq X_n\mid |S|=2k-1, 1\le k\le n\}.$ Therefore $$|\mathcal E_n|=\sum_{k=0}^n\binom{3n}{2k}\quad\&\quad |\mathcal O_n|=\sum_{k=1}^n\binom{3n}{2k-1}.$$
Let $\begin{aligned}\mathcal A_n&:=\{S\in\mathcal E_n\mid |S|=2n,1\notin S\}\\\mathcal B_n&:=\{S\in\mathcal E_n\mid |S|=2k,0\le k\le n-1,1\notin S\}\\\mathcal C_n&:=\{S\in \mathcal E_n\mid 1\in S\}\end{aligned}$
and $\begin{aligned}\mathcal P_n&:=\{S\in\mathcal O_n\mid 1\in S\}\\\mathcal Q_n&:=\{S\in \mathcal O_n\mid 1\notin S\}.\end{aligned}$.
All these $5$ collections are mutually disjoint and we have $|\mathcal E_n|=|\mathcal A_n|+|\mathcal B_n|+|\mathcal C_n|$ and $|\mathcal O_n|=|\mathcal P_n|+|\mathcal Q_n|$
You can continue by claiming that $|\mathcal B_n|=|\mathcal P_n|$ and $|\mathcal C_n|=|\mathcal Q_n|.$
For the first claim, $\forall S\in\mathcal B_n,|S|\le 2n-2,|S|$ is even, so $|S\cup\{1\}|=|S|+1\le 2n-1$ and $|S\cup\{1\}|$ is odd. $\implies S\cup\{1\}\in\mathcal P_n.$ $f:\mathcal B_n\to\mathcal P_n, f:S\mapsto A\cup \{1\}$ is a bijection $\implies |\mathcal B_n|=|\mathcal P_n|.$
Similarly for $|\mathcal C_n|=|\mathcal Q_n|.$
Therefore: $|\mathcal E_n|=|\mathcal A_n|+|\mathcal B_n|+|\mathcal C_n|=|\mathcal A_n|+|\mathcal P_n|+|\mathcal Q_n|=|\mathcal A_n|+|\mathcal O_n|.$
Finally, we obtain $$\begin{aligned}\sum_{k=0}^n\binom{3n}{2k}&=\binom{3n-1}{2n}+\sum_{k=1}^n\binom{3n}{2k-1}\\\implies \sum_{k=0}^{2n}(-1)^k\binom{3n}k&=\binom{3n-1}{2n}.\end{aligned}$$
