Question

How many 4 digit numbers are there which contain not more than 2 different digits ?

Answer 1

Case 1 : When all digits are same : ${}^9{C}_1$ (excluding $0$)

Case 2: When digits are different and $0$ excluded.

(a) Selecting 2 numbers in ${}^9{C}_2$ ways
(b) Each digit can be filled in 2 ways hence $2222=24$ way
(c) Undesirable case : when a particular digit is same $(21=2$ ways)

(case 1)

$\therefore {}^9{C}_2(2^4-2)$ ways

Case 3: When digits are different and $0$ is included

(a) other digit can be chosen in ${}^9{C}_1$ ways

(b) 0 can't be placed at ten thousand's place, hence the selected digit should be fixed at this place remaining 3 digits can be filled with $222=2^3$ ways.

(c) Undesirable case: When all the 4 digits gets filled with selected digit only (0 not included)= 1 way hence no. of ways will be: ${}^9{C}_1(2^3-1)$

$\therefore$ Total desired ways :- ${}^9{C}_1+{}^9{C}_2(2^4-2)+{}^9{C}_1(2^3-1)=576$ ways