Question

How many numbers each lying between $9$ and $1000$ can be formed with the digits $0,1,2,3,7,8$ if repetition of the digits in a number is allowed?

Answer 1

List of the digits that are allowed = $0,1,2,3,7,8$

Repetition is allowed

Two digit number:

First digit has $5$ Ways $(1,2,3,7,8)$

Second digit has $6$ Ways $(0,1,2,3,7,8)$

Number of possible two digit number is $30$

Three digit number:

First digit has $5$ Ways $(1,2,3,7,8)$

Second digit has $6$ Ways $(0,1,2,3,7,8)$

Third digit has $6$ Ways $(0,1,2,3,7,8)$

Number of possible three digit number is $180$

Total number possible from $9$ to $1000$

$=180+30$

$=210$ Numbers