Question

Write the greatest $6$-digit number using only $0,1,8$ and $9$ digits. All the digits must be used at least once.

Answer 1

Hint: We need to find greatest number using digits $0,1,8$ and $9$
Here we have $4$-digits and find greatest $6$-digit number
Therefore we have to arrange digits in repeatedly manner
When to find the greatest digit we have to arrange digits in descending order.
By arranging the digits in descending order, we are arranging some digits in a repeated manner.
By doing some arrangements the required result will achieve.
And then we will get the required answer.

Complete step-by-step answer:
It is given that the digits $0,1,8$ and $9$
We need to find the greatest \[6\]-digit number using the above given digits.
Here we have $4$- different digits but we have to find greatest $6$-digit number
We have approach to get the greatest $6$-digit number with have only $4$-digits
When we write the greatest number we need to arrange digits in decreasing order.
Here given digits are lesser than the desired number.
So we need to place digits in repeatedly manner in descending order
We have to place $9$ up to $3$ places from left end of the number
It looks like $999$
After that decreasing number is \[8\] so we get the number is $9998$
By adding remaining $2$-digits (which are$1,0$), we will arrange them in descending order
By arranging the remaining number in repeatedly manner is that manner is $999810$

Hence we get the greatest $6$-digit number from the digits $0,1,8$ and $9$ is $999810$

Note: We should know that the n-digit number has a total number of digits equal to n.
If we do not repeat digits here we could not get the $6$-digit greatest number from given $4$-digits.
For the greatest number we need to arrange digits in descending order and for smallest numbers we need to arrange digits in ascending order.