The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4, (repetition of digits is not allowed) and are multiples of 3 is?
Solutions · 1
Official solution
Here number should be divisible by 3, that means sum of numbers should be divisible by 3.
Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are
(1) (\,\,\,\,\] (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)
(2) \[\,\,\,\,\] (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3 = 6 (divisible by 3)
Case 1 :
When 4 digits are (0, 2, 3, 4) then
\[\therefore\,\,\,\,\] Total possible numbers = 3C1 \[ \times \] 3C1 \[ \times \] 2C1 \[ \times \] 1C1
= 3 \[ \times \] 3 \[ \times \] 2 \[ \times \] 1 = 18
Case 2 :
When 4 digits are (0, 1, 2, 3) then,
\[\therefore\,\,\,\,\] Total possible number in this case = 2C1 \[ \times \] 3C1 \[ \times \] 2C1 \[ \times \] 1C1
= 2 \[ \times \] 3 \[ \times \] 2 \[ \times \] 1 = 12
\[\therefore\,\,\,\,\] Total possible numbers will be = 18 + 12 = 30