Fundamental Drill#4

From the first two constraints, we have $0\le x \le 1009$ . Then $\dbinom {2018}x \ge \dbinom {2018}{x+1009} = \dbinom {2018}{2018-(x+1009)} = \dbinom {2018}{1009-x}$

For $0 \le x \le 1009$ , $\dbinom {2018}{\color{#3D99F6}x} \ge \dbinom {2018}{\color{#3D99F6}1009-x}$ if $\color{#3D99F6}x \ge 1009-x$ , $\implies 2x \ge 1009$ , $\implies x \ge 505$ . Therefore, $505 \le x \le 1009$ , altogether $\boxed{505}$ possible values.

$\begin{cases} x\geq0 \\ x+1009\leq2018 \\ \binom{2018}{x}\geq \binom{2018}{x+1009} \end{cases}$

The first two inequalities give us $0\leq x\leq1009$

$\binom{2018}{x+1009}=\binom{2018}{1009-x}$

See the proof here .

$\Rightarrow \binom{2018}{x}\geq \binom{2018}{1009-x}$

When $x=0$ , $\binom{2018}{x}=\binom{2018}{0}, \binom{2018}{1009-x}=\binom{2018}{1009}$

When $x=1$ , $\binom{2018}{x}=\binom{2018}{1}, \binom{2018}{1009-x}=\binom{2018}{1008}$

When $x=2$ , $\binom{2018}{x}=\binom{2018}{2}, \binom{2018}{1009-x}=\binom{2018}{1007}$

...

When $x=504$ , $\binom{2018}{x}=\binom{2018}{504}, \binom{2018}{1009-x}=\binom{2018}{505}$

When $x=505$ , $\binom{2018}{x}=\binom{2018}{505}, \binom{2018}{1009-x}=\binom{2018}{504}$

...

When $x=1007$ , $\binom{2018}{x}=\binom{2018}{1007}, \binom{2018}{1009-x}=\binom{2018}{2}$

When $x=1008$ , $\binom{2018}{x}=\binom{2018}{1008}, \binom{2018}{1009-x}=\binom{2018}{1}$

When $x=1009$ , $\binom{2018}{x}=\binom{2018}{1009}, \binom{2018}{1009-x}=\binom{2018}{0}$

Notice that the values of both sides are interchanged throughout the process of substituting values.

From $x=0$ to $x=1009$ , there are a total of $1009+1=1010$ values

$\therefore$ the number of possible values for $x=\cfrac{1010}{2}=\boxed{505}$ ,

which starts from $x=505$ all the way to $x=1009$