Different digit

Two avoid double-usage of the term 'digit', I'll use 'character' for the symbol and 'digit' for the position.

Consider three cases:

1) There are $9$ four-digit numbers that use only one character: $1111,2222,\ldots,9999$ .

2) Now let's count four-digit numbers that have two different characters where neither character is $'0'$ .

There are ${9 \choose 2}$ ways to choose two different characters. There are $2^4$ strings of length $4$ composed of the two characters, but we must remove the two constant strings from consideration. Thus there are $2^4 - 2$ ways to choose the digits for the two characters so that each is used at least once.

Thus there are ${9 \choose 2} * (2^4 -2) = 36*14 = 504$ four digit numbers that use exactly two different characters.

3) Now we consider four-digit numbers that include the $'0'$ character. There are $9$ choices for the other character (which appears in the thousands position).
We must choose a non-empty subset of the other three positions to hold zeros. There are $2^3 -1 = 7$ such subsets of the three other positions.
Thus there are $9*7=63$ four-digit numbers that contain exactly two different characters, one of which is $'0'$ .

The total number of four-digit numbers with at most two different characters is thus $9+504+63 = 576$ .

Let

• $a$ be ${1,2,3,4,5,6,7,8,9}$
• $b$ be ${0,1,2,3,4,5,6,7,8,9}$ but $\neq a$

There are 7 cases with two different digits:

$abbb, aabb, abab, abba, aaab, aaba, abaa$

each got $9\times9 = 81$ choices , total $= 7 \times 81 = 567$

for single digit number $aaaa$ , there are $9$ choices.

So total number of 4-digit numbers which contain not more than 2 different digits $= 567 + 9 = \boxed{576}$

from 1 to 9 there are 9 digits, first chose 2 out of these in 9C2 ways Now all the four places can be filled in 16 ways 16*9C2 = 576