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by sanju09 » Thu Jan 28, 2010 12:38 am
A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. What is the total number of ways in which this can be done?
(A) 216
(B) 240
(C) 600
(D) 3125
(E) 6250
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by ajith » Thu Jan 28, 2010 12:48 am
sanju09 wrote:A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. What is the total number of ways in which this can be done?
(A) 216
(B) 240
(C) 600
(D) 3125
(E) 6250
For a number to be divisible by 3 the sum of digits should be divisible by 3

from 0,1,2,3,4,5 there are 6 ways of choosing 5 digits (6C1)

of which 0,1,2,4,5 and 1,2,3,4,5 are the only combination which has sum divisible by 3

Now with 1,2,3,4,5 we can create 5! numbers that is 120

With 0,1,2,4,5 since 0 cant be the first number we can have only 4*4! = 96

The total comes to 126+96 = 216
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by money9111 » Thu Jan 28, 2010 2:34 am
i think i would have gotten as far as 96 and wouldn't have gone that extra step to think.. ok now what... i would have looked at the answer choices and said "OH WTF"...then started panicking... then guessed... woe is me
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