First slide

Permutations

Question

Let $a_{n} = \frac{10^{n}}{n!}$ for $r n \geq 1 .$ Then $a_{n}$ take the greatest value when $n$ equals

Moderate

A

20
B

18
C

6
D

9

Solution

We have $\frac{a_{n}}{a_{n + 1}} = \frac{10^{n}}{n!} \times \frac{(n + 1)!}{10^{n + 1}} = \frac{n + 1}{10}$
Note that $\frac{a_{1}}{a_{2}} < 1, \frac{a_{2}}{a_{3}} < 1, \frac{a_{3}}{a_{4}} < 1, \frac{a_{4}}{a_{5}} < 1$
$\dots \frac{a_{8}}{a_{9}} < 1, \frac{a_{9}}{a_{10}} = 1, \frac{a_{10}}{a_{11}} > 1, \frac{a_{11}}{a_{12}} > 1, \dots$
Thus, $a_{1} < a_{2} < a_{3} < \dots < a_{9} = a_{10} > a_{11} > a_{12}$
$∴ a_{n}$ is greatest if $n = 9 or 10 .$

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