Permutations and Combinations Part 1

Given, digits are =1,2,2,3
∴ Total distinct numbers =4!∕2!=12
1 at unit place ⇒ Number of such numbers =‌

3!

2!

=3
2 at unit place ⇒ Number of such numbers =3!=6
3 at unit place ⇒ Number of such numbers =‌

3!

2!

=3
∴ Sum of digits at unit place is
3×1+6×2+3×3=24
Hence, sum of all 4 digit such numbers
=(3+12+9)(103+102+10+1)
=1111×24
=26664