**Quote:**

How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated?

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Choices

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A. 15

B. 96

C. 216

D. 120

E. 625

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Case 1: Any 5-digit number formed using the numbers in the set {1,2,3,4,5} is divisible by 3 because 1+2+3+4+5=15 which is divisible by 3.

Thus, number of 5-digit numbers = 5! = 120

Case 2: Any 5-digit number formed using the numbers in the set {0,1,2,4,5} is divisible by 3.

Thus, number of 5-digit numbers = 5! - 4! = 96

4! is subtracted to remove all the permutations where 0 comes as the first number. For example, 01254 is not a 5-digit number. It becomes a 4-digit number. Hence, with 0 as the first number, there are 4! permutations. This should be deducted from 5!

Case 3: No other combination of 5 numbers will be divisible by 3.

From Cases 1, 2 and 3, number of 5-digit numbers = 120 + 96 = 216. (C)

Note that they have asked only for positive integers. If not, we should do 216*2
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