Question

How many 4digit no. are there which contains not more than 2 different digits?

Answer 1

The first (non-zero) digit of the number FF(thousands digit) can be any one of nine. The second digit SS which is used if there are two digits can be any one of the nine digits different from the first.

Now consider the Hundreds Tens and Units digits in the case that there are two digits used in the number. We have two possibilities FF or SS to fill each place - but we exclude FFFFFF as not involving two digits, so there are
${2}^{3} - 1 = 7$
possible patterns with exactly two different digits and 9×9 ways of choosing the pair
of digits in first place
Then there are nine possibilities with just one digit
so the total is
9×9×7+9
=576