If f(n) = n! (31-n)!,where n in {0, 1, 2, ldo | vedclass

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(D) Given $f(n) = n! (31-n)!$. We observe that $f(n) = \frac{31!}{\binom{31}{n}}$. To minimize $f(n)$,we need to maximize the binomial coefficient $\b

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$(15!) (16!)$

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(D) Given $f(n) = n! (31-n)!$. We observe that $f(n) = \frac{31!}{\binom{31}{n}}$. To minimize $f(n)$,we need to maximize the binomial coefficient $\binom{31}{n}$. The binomial coefficient $\binom{N}{n}$ is maximized when $n = \lfloor N/2 \rfloor$ or $n = \lceil N/2 \rceil$. For $N = 31$,the maximum values occur at $n = 15$ and $n = 16$. Thus,the minimum value of $f(n)$ is $f(15) = 15! (31-15)! = 15! 16!$. Similarly,$f(16) = 16! (31-16)! = 16! 15!$.

If $f(n) = n! (31-n)!$,where $n \in \{0, 1, 2, \ldots, 31\}$,then the minimum value of $f(n)$ is

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