JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 1)
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 multiple of 3 is :
24
30
36
48
Explanation
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
_16th_April_Morning_Slot_en_1_1.png)
$$\therefore\,\,\,\,$$ 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,
_16th_April_Morning_Slot_en_1_2.png)
$$\therefore\,\,\,\,$$ 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
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
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