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

  • A
    $(15!) (15!)$
  • B
    $(15!) (14!)$
  • C
    $(14!) (16!)$
  • D
    $(15!) (16!)$

Explore More

Similar Questions

How many $5$-digit numbers can be formed using the digits $0, 1, 2, 3, 4,$ and $5$ without repetition such that the number is divisible by $3$?

Difficult
View Solution

For which value of $n \in N$,does $n!$ have $13$ trailing zeros?

If ${}^n C_{r-1}=36$, ${}^n C_r=84$, and ${}^n C_{r+1}=126$, then the value of ${}^n C_8$ is

If all the letters of the word $COMBINATION$ are arranged in all possible ways to form $11$ letter words (with or without meaning),then the number of words among them in which $C$ and $N$ occupy the end positions and no vowel appears exactly in the middle position is

$A$ number is called a palindrome if it reads the same backward as well as forward. For example,$285582$ is a six-digit palindrome. The number of six-digit palindromes,which are divisible by $55$,is ...... .

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo