Question
The total number of ways in which a 5 digit number divisible by 3 can be formed from the digits 0,1,2,3,4,5 without repetition is?
Answer: Option C
:
C
option (c )
Divisibility test for 3- sum of digits should be divisible by 3.
Case 1:- When 0 is not included → sum = 15 divisible by 3
Number of combinations = 5! = 120
Case 2:- When 3 is not used. → Sum = 12 divisible by 3
Number of combinations - 4.4..3.2 =96
Total = 120+96= 216
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:
C
option (c )
Divisibility test for 3- sum of digits should be divisible by 3.
Case 1:- When 0 is not included → sum = 15 divisible by 3
Number of combinations = 5! = 120
Case 2:- When 3 is not used. → Sum = 12 divisible by 3
Number of combinations - 4.4..3.2 =96
Total = 120+96= 216
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