QJEEM18Mae69d: JEE Main 2018 Mathematics - Class 11 JEEM18Mae69d Solution: The number of numbers between 2,000 and 5,000 that can be…

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 = $^3{C_1}$ $\times$ $^3{C_1}$ $\times$ $^2{C_1}$ $\times$ $^1{C_1}$

=    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 = $^2{C_1}$ $\times$ $^3{C_1}$ $\times$ $^2{C_1}$ $\times$ $^1{C_1}$

=    2 $\times$ 3 $\times$ 2 $\times$ 1 = 12

$\therefore\,\,\,\,$ Total possible numbers will be = 18 + 12 = 30