How many four-digit numbers are there whose decimal notation contains not more than two distinct digits?

Explanation

We have to form four-digit numbers whose decimal notation contains one or two different digits. Clearly the number of four-digit numbers with one digit such as 1111, 2222)=9 etc.
Let us consider two different digits
Here two cases arise.
Case I :
Out of two digits, one digits is 0. We can select other non-zero digit in `""^(9)C_(1)` =0 ways. Let us consider the digits 1,0 . Then the four digits numbers are 1110, 1100, 1000 and their arrangements . When thousand place is 1, the remaining places can be arranged in `((3!)/(2!)+(3!)/(2!)+1)=7` ways
`therefore` The number of required number `=7xx9=63`
Case II
Two digits are non-two different digits. We can select two different non-zero digit out of 9 non zero digits in `""^(9)C_(2)` = 36 ways .
Let us proceed with 1,2

Then possible combinations for four digitts are 1,1,2,2,1,1,1,2 and 2,2,2,1 and their corresponding arrangements are `(4!)/((2)!(2)!), (4!)/(3!) and (4!)/(3!)` respectively .
Hence the number of four digits numbers under given condition
`=""^(9)C_(2)(6+4+4)=36xx14=504`
`therefore` Total number of four-digit numbers whose decimal notation contains not more than two distinct digits.
`=9+63+504=576`