If $\mathrm{f}(\mathrm{n})=\mathrm{n} !(31-\mathrm{n}) !$, where $\mathrm{n} \in\{0,1,2, \ldots, 31\}$, then…
If $\mathrm{f}(\mathrm{n})=\mathrm{n} !(31-\mathrm{n}) !$, where $\mathrm{n} \in\{0,1,2, \ldots, 31\}$, then the minimum value of $f(n)$ is
- $(15!) (15!)$
- $(15 !)(14 !)$
- $(14!) (16!)$
- $(15 !)(16 !)$
Solution
$\begin{aligned} & \text {} \because f(n)=n !(31-n) ! \\ & \text { So, } f(31-n)=(31-n !)(n !)=f(n) \Rightarrow f(0)=f(31) \\ & f(1)=f(30) \ldots \\ & \text { Also } f(0)>f(1)>f(2)>\ldots>f(15) < f(16) < f(17) \ldots \\ & < f(30) < f(31) \\ & \therefore f(15) \text { is minimum. Then, the minimum value is: } \\ & \Rightarrow f(15 !)=15 !(31-15) !=(15 !)(16 !)\end{aligned}$
Asked in: AP EAMCET 2023 (17 May Shift 1)
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