Question

If a $5$ digit number is made using all the digits from $1,3,4,6,8$, such that all the digits of number from $1$st position to $5$th should not be in increasing order, then the position of number ${}^{\prime \prime }{63184}^{\prime \prime }$ after listing all the numbers formed in ascending order is

• A. $77$
• B. $78$
• C. $79$
• D. $80$

Answer 1

The correct option is C $79$

Digits available are $1,3,4,6,8$
Since the numbers had to be arranged in ascending order.
Total number starting from $1$ $\left(1\_\_\_\_\right)$ will be $4!$
Total number starting from $3$ $\left(3\_\_\_\_\right)$ will be $4!$
Total number starting from $4$ $\left(4\_\_\_\_\right)$will be $4!$
Now the total numbers starting from $61$$\left(61\_\_\_\right)$ will be $3!$
Now the total numbers starting from $6314$ $\left(6314\_\right)$will be $1$ only
Now the next number will be ${}^{\prime \prime }{63184}^{\prime \prime }$

Also the digits of the number should not be in increasing order, here only one number will have digits in increasing order $13468$.

Total number before ${}^{\prime \prime }{63184}^{\prime \prime }$ are $4!+4!+4!+3!+1-1=78$
$\therefore$ The Required position of number ${}^{\prime \prime }{63184}^{\prime \prime }$ after arranging all number in ascending order will be $79$