We have,
$A_n =\max \left\{{ }^n C_r \mid 0 \leq r \leq n\right\}$
$n \in\{1,2,3, \ldots, 20\}$
Case $I$ When $n$ is even
$A_n={ }^n C_{n / 2}$
$\therefore \quad \frac{A_n}{A_{n-1}}=\frac{{ }^n C_{n ' 2}}{{ }^{n-1} C_{\frac{n-1-1}{}}^2}=2$
So for all $n$ even given relation is true.
Case II When $n$ is odd
$A_n={ }^n C_{\frac{n-1}{}}^2$
$\therefore \quad \frac{A_n}{A_{n-1}} =\frac{{ }^n C_{n-1}}{{ }^{n-1} C_{n-1}}=\frac{2 n}{n+1}$
$19 \leq \frac{2 n}{n+1} \leq n \text { if } n=19$
$\therefore$ Total number of elements are 10 even number and $19=10+1=11$
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