How many four digit numbers divisible by 9 can be formed using the digits { 0,1,2,3} ?

Question

{3, 3, 2, 1} --> 4!=24

{2, 2, 2, 3}  -->4!=24

{3, 3, 3, 0)  --> 4!-3!= 18

24+24+18=66 But the correct answer is 62. Why?

Answer 1

The correct answer is not 62, it is 19.

{3, 3, 2, 1} has 12 arrangements C(4,2) spots for the 3s, then 2! spots for the 2 and 1. 6 * 2 = 12

{2, 2, 2, 3} has 4 arrangements - the 3 can be in 4 different spots.

{3, 3, 3, 0} has 3 arrangements with 0 not as the first digit - the 0 can be in 3 different spots.