For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$

Question

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.

Answer 1

It is even true for $n\ge3$. Indeed $\;\dbinom 93=\dfrac{9!}{3!\,6!}=84 >64$.

\begin{align} &\text{Next, we have }\qquad\qquad &\binom{3(n+1)}{n+1}&=\binom{3n}n\,\frac{(3n+1)(3n+2)(3n+3)}{(n+1)(2n+1)(2n+2)},&&\qquad\qquad\end{align} so, by induction, all we have to prove is that $$\frac{3(3n+1)(3n+2)}{2(n+1)(2n+1)}\ge4$$ This inequality is equivalent to $$ 27n^2+27n+6\ge 16n^2+24n+8\iff 11n^2+3n\ge 2, $$ which is obviously true if $n \ge 1$.

Answer 2

Your thinking seems correct to me. Here's a similar approach: The cases $n=4,5$ can be checked by hand. For $n\geq6$ we have $n!\leq\left(\frac{n}{2}\right)^n$, which is easily proved by induction, and hence $$\binom{3n}{n}=\frac{(3n)!}{(2n)!n!}=\prod_{i=k}^n\frac{2n+k}{k}=\prod_{k=1}^n\left(1+\frac{2n}{k}\right)>2^n\frac{n^n}{n!}\geq4^n,$$ where the first inequality comes from the fact that $1+\frac{2n}{k}>\frac{2n}{k}$ for all $k$.

Answer 3

We have the following Stirling approximations:

• $\sqrt{2\pi n}n^ne^{-n} \leq n! \leq e \sqrt{n} n^n e^{-n} $ for all $n \in \mathbb{N}$ $${{3n} \choose {n}} \geq \frac{\sqrt{2\pi 3n}(3n)^{3n}e^{-3n}}{e \sqrt{n} n^n e^{-n} \cdot e \sqrt{2n} (2n)^{2n} e^{-2n}} =\frac{\sqrt{3\pi}}{e^2\sqrt{n}} \frac{27^n}{4^n} > \frac{\sqrt{3\pi}}{e^2} \frac{\left(\frac{3}{2}\right)^n}{\sqrt{n}}4^n$$ So, if $n$ is large enough, the inequality holds.

Answer 4

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

${ {3n} \choose {n}} \ge 3^n + \sum_{k=2}^n { n \choose {n-k}}3^{n-k} A_k\ge \sum_{k=0}^n { n \choose {n-k}}3^{n-k} =(3+1)^n =4^n$