Question

If a $6$ digit number is made using all the digits $1,2,4,5,7,8$, then the position of number ${}^{\prime \prime }{541782}^{\prime \prime }$ when all numbers formed are arranged in ascending order is

Answer 1

Available digits: $1,2,4,5,7,8$
Since the numbers had to be arranged in increasing order.
Total number starting from $1$ $\left(1\_\_\_\_\_\right)$ will be $5!$
Total number starting from $2$ $\left(2\_\_\_\_\_\right)$ will be $5!$
Total number starting from $4$ $\left(4\_\_\_\_\_\right)$ will be $5!$
Now the total numbers starting from $51$ $\left(51\_\_\_\_\right)$ will be $4!$
Total numbers starting from $52$ $\left(52\_\_\_\_\right)$ will be $4!$
Now the total numbers starting from $5412$ $\left(5412\_\_\right)$ will be $2!$
Now the total numbers starting from $54172$ $\left(54172\_\right)$ will be $1$ only
Now the next number will be ${}^{\prime \prime }{541782}^{\prime \prime }$
Total number before ${}^{\prime \prime }{541782}^{\prime \prime }$ are $5!+5!+5!+4!+4!+2!+1=411$
$\therefore$ The position of ${}^{\prime \prime }{541782}^{\prime \prime }$ in increasing order will be $412$