For each positive integer n,let A_n = max lef | vedclass

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(C) We have $A_n = \max \left\{ \binom{n}{r} \mid 0 \leq r \leq n \right\}$.Case $I$: When $n$ is even,$A_n = \binom{n}{n/2}$.Then $\frac{A_n}{A_{n-1}

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$11$

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(C) We have $A_n = \max \left\{ \binom{n}{r} \mid 0 \leq r \leq n ight\}$.Case $I$: When $n$ is even,$A_n = \binom{n}{n/2}$.Then $\frac{A_n}{A_{n-1}} = \frac{\binom{n}{n/2}}{\binom{n-1}{(n-2)/2}} = \frac{n!}{(n/2)!(n/2)!} 	imes \frac{((n-2)/2)!(n/2)!}{(n-1)!} = \frac{n}{n/2} = 2$.Since $1.9 \leq 2 \leq 2$,all even values of $n$ in the set $\{1, 2, \ldots, 20\}$ satisfy the condition. There are $10$ such values $(n = 2, 4, \ldots, 20)$.Case $II$: When $n$ is odd,$A_n = \binom{n}{(n-1)/2} = \binom{n}{(n+1)/2}$.Then $\frac{A_n}{A_{n-1}} = \frac{\binom{n}{(n-1)/2}}{\binom{n-1}{(n-1)/2}} = \frac{n!}{(\frac{n-1}{2})!(\frac{n+1}{2})!} 	imes \frac{(\frac{n-1}{2})!(\frac{n-1}{2})!}{(n-1)!} = \frac{n}{(n+1)/2} = \frac{2n}{n+1}$.We require $1.9 \leq \frac{2n}{n+1} \leq 2$.The inequality $\frac{2n}{n+1} \leq 2$ is always true for $n > 0$ as $2n \leq 2n+2$.The inequality $1.9 \leq \frac{2n}{n+1}$ implies $1.9n + 1.9 \leq 2n$,so $0.1n \geq 1.9$,which means $n \geq 19$.For $n \in \{1, 3, \ldots, 19\}$,only $n = 19$ satisfies this condition.Total values of $n$ are $10$ (even) $+ 1$ (odd,$n=19$) $= 11$.

For each positive integer $n$,let $A_n = \max \left\{ \binom{n}{r} \mid 0 \leq r \leq n \right\}$. Then,the number of elements $n \in \{1, 2, \ldots, 20\}$ for which $1.9 \leq \frac{A_n}{A_{n-1}} \leq 2$ is

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