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Problem 169

AMC 10/12, early questions

Combinatorics Difficulty 3.6 Find the answer

Given that $C_{x}^{m}=\frac{x(x-1)…(x-m+1)}{m!}$ , where $x\in R$ , $m$ is a positive integer, this is an extension of the combination number $C_{n}^{m}$ ( $m$ , $n$ are positive integers, and $m\leqslant n$ ). If $x > 0$ , find the minimum value of $\frac{C_{x}^{3}}{(C_{x}^{1})^{2}}$ .

Your answer

Official solution

According to the given conditions, when $x > 0$ ,
$\frac{C_{x}^{3}}{(C_{x}^{1})^{2}}=\frac{x(x-1)(x-2)}{6x^{2}}$
$=\frac{x^{2}-3x+2}{6x}$
$=\frac{x}{6}+\frac{1}{3x}-\frac{1}{2}$
By the AM-GM inequality, we have
$\frac{x}{6}+\frac{1}{3x}-\frac{1}{2}\geqslant 2\sqrt{\frac{x}{6}\cdot\frac{1}{3x}}-\frac{1}{2}$
$=\frac{\sqrt{2}}{3}-\frac{1}{2}$
Equality holds if and only if $\frac{x}{6}=\frac{1}{3x}$ , which is when $x=\sqrt{2}$ .

Therefore, the minimum value of $\frac{C_{x}^{3}}{(C_{x}^{1})^{2}}$ is $\boxed{\frac{\sqrt{2}}{3}-\frac{1}{2}}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.