How many odd numbers less than $1000$ can be formed by using the digits $0, \; 3, \; 5, \; 7$ when repetition of digits is not allowed?
The required numbers are 3 digit OR 2 digit OR 1 digit odd numbers.
Case 1: 3-digit odd numbers
The unit's place can be selected from the digits $3, \; 5, \;$ and $7$ in $3$ ways.
The ten's place can either be a 0 or can be selected from the remaining odd digits.
If the ten's place is 0, it can be selected in $1$ way.
OR, the ten's place can be selected from remaining two odd digits in $2$ ways.
When the ten's place is 0, then the hundred's place can be selected from the remaining two odd digits in $2$ ways.
OR, when the ten's place is an odd digit, then the hundred's place can be selected from the remaining odd digit in $1$ way.
$\therefore \;$ Number of ways of forming a 3-digit odd number less than $1000$ using the given digits is
$= 3 \times 1 \times 2 + 3 \times 2 \times 1 = 12$ ways
Case 2: 2-digit odd numbers
The unit's place can be selected from the digits $3, \; 5, \;$ and $7$ in $3$ ways.
The ten's place can be selected from the remaining two odd digits in $2$ ways.
$\therefore \;$ Number of ways of forming a 3-digit odd number less than $1000$ using the given digits is $= 3 \times 2 = 6$ ways
Case 3: 1-digit odd numbers
Number of ways of forming 1-digit odd numbers from the given digits is $3$ ways
$\therefore \;$ Total number ways of forming odd numbers less than $1000$ by using the given digits is $12 + 6 + 3 = 21$ ways
i.e. $\;$ $21$ odd numbers can be formed from the given digits which are less than 1000.