<s>[INST] For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes.  Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$

By induction:  $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$.  Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{n}\right){{2n-1}\choose{n-1}} > 2\left(\frac{2n-1}{n}\right)\frac{4^{n-1}}{n-1} > 2 \cdot 2\cdot \frac{4^{n-1}}{n} = \frac{4^n}{n}$
Note:  This argument was taken from Wikipedia.

(2)  $(3n-1) > \left(\frac{3}{2}\right)(2n-1), (3n-2) > \left(\frac{3}{2}\right)(2n-2), \dots, (2n-n+1) > \left(\frac{3}{2}\right)(2n-n+1)$

If $2a = 3b$, then $2a - 2 > 3b - 3$ and $2(a-1) > 3(b-1)$ so that $(a-1) > \frac{3}{2}(b-1)$

(3)  For $n\ge 2$, ${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n} \choose {n}}$

From (2) above: ${{3n} \choose{n}} = \frac{(3n)(3n-1)\dots(3n-n+2)(3n-n+1)}{n!} > \left(\frac{3}{2}^n\right)\frac{(2n)(2n-1)\dots(2n-n+2)(2n-n+1)}{n!} = \left(\frac{3}{2}^n\right){{2n} \choose {n}}$

(4)  Since for $n \ge 4, \left(\frac{3}{2}\right)^n > n$, it follows that:

${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n}\choose{n}} > n\left(\frac{4^n}{n}\right) = 4^n $
Note: The argument that for $n \ge 2, \left(\frac{3}{2}\right)^n > n$ can be found here.

 [/INST] A combinatorial argument
Divide $3n$ distinctive elements into $n$ groups, each of which contains 3 elements. To pick $n$ elments out of them, there are many ways, including the following categories:
1) pick one element out of each group, which gives $3^{n}$ different ways
2) pick one element out of each group for $n-k$ group, then pick $k$ elments out of remaining $k$ groups, so that every group of the remaining $k$ groups contributes either $0$ elements or at least $2$ elements. This gives ${ n \choose {n-k}}3^{n-k} A_k$ ways, where $A_k$ is the way of picking $k$ elments out of remaining $k$ groups, so that every group of the remaining $k$ groups contributes either $0$ elements or at least $2$ elements. Obviously $A_k \ge 1$, except for $k=1$we have $A_1=0$
Remark that the above ways are exlcusive for different $k$.
With ${ n \choose {n-2}}3^{n-2} A_2 = { n \choose {n-2}}3^{n-2} 6 = n(n-1)3^{n-1}\ge 3^{n-1} n + \frac{n(n-1)}{2}3^{n-2}$ (the last inequality is equivalent to $5n-11\ge 0$)
we have
${ {3n} \choose {n}} \ge 3^n + \sum_{k=2}^n { n \choose {n-k}}3^{n-k} A_k\ge \sum_{k=0}^n { n \choose {n-k}}3^{n-k} =(3+1)^n =4^n$
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