Question

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

• A. \( (15!)(15!) \)
• B. \( (15!)(14!) \)
• C. \( (14!)(16!) \)
• D. \( (15!)(16!) \)

Hint:

To minimize a product of two factorials, balance the two terms by choosing a value of \( n \) close to the midpoint.

Answer 1

The Correct Option is D

We are given the function \( f(n) = n!(31-n)! \). To minimize this function, we need to find the value of \( n \) that minimizes the product of the two factorials. This occurs when \( n = 15 \), as the product of the factorials \( 15! \) and \( 16! \) will yield the smallest value. Thus, the minimum value of \( f(n) \) is \( (15!)(16!) \).