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 descending order is

Answer 1

Available digits: $1,2,4,5,7,8$
Since the numbers had to be arranged in decreasing order.
Total number starting from $8$ $\left(8\_\_\_\_\_\right)$ will be $5!$
Total number starting from $7$ $\left(7\_\_\_\_\_\right)$ will be $5!$
Now the total numbers starting from $58$ $\left(58\_\_\_\_\right)$ will be $4!$
Total numbers starting from $57$ $\left(57\_\_\_\_\right)$ will be $4!$
Now the total numbers starting from $548$ $\left(548\_\_\_\right)$ will be $3!$
The total numbers starting from $547$ $\left(547\_\_\_\right)$ will be $3!$
The total numbers starting from $542$ $\left(542\_\_\_\right)$ will be $3!$
Now the total numbers starting from $5418$ $\left(5418\_\_\right)$ will be $2!$
Now the next number will be ${}^{\prime \prime }{541782}^{\prime \prime }$
Total number before ${}^{\prime \prime }{541782}^{\prime \prime }$ are $5!+5!+4!+4!+3!+3!+3!+2!=308$
$\therefore$ The required position of ${}^{\prime \prime }{541782}^{\prime \prime }$ will be $309$