Wipro
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Numerical Ability
Permutation and Combination
How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)
A. 216
B. 3152
C. 240
D. 600
E. 305
Read Solution (Total 2)
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- ans is:-option A 216
- 10 years agoHelpfull: Yes(0) No(0)
- Here,
according to the question the five digit number should be divisible by 3.......
(wkt,sum of the digits should be divisible by 3)
so only possible 5 digit numbers are{1,2,3,4,5} => sum is 15 and {0,1,2,4,5} sum=>12
{1, 2, 3, 4, 5} --> 5! =120
and for {0,1,2,4,5}
we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96
120+96=216 is the answer
- 10 years agoHelpfull: Yes(0) No(0)
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