A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE DIGITS 0, 1, 2, 3, 4 and 5 WITHOUT REPETITIONS. THAT TOTAL NO OF WAYS IT CAN BE DONE IS?
Select.
A) 122  B) 210  C) 216  D) 217  E) 220
First step:   We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.   We have six digits: 0,1,2,3,4,5. Their sum=15.   For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0=(1,2,3,4,5) and 15-3=(0,1,2,4,5). Meaning that no other 5 from given six will total the number divisible by 3.   Second step:   We have two set of numbers:   1,2,3,4,5 and 0,1,2,4,5. How many 5 digit numbers can be formed using this two sets:   1,2,3,4,5 --&gt; 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.   0,1,2,4,5 --&gt; here 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! --&gt; 5!-4!=96   120+96= 
<gadget id="calculator">120+96</gadget>
<output>216</output>
216   Answer: 216.
<result>C</result>
A FIVE DIGIT NUMBER DIVISIBLE BY 3 IS TO BE FORMED USING THE DIGITS 0, 1, 2, 3, 4 and 5 WITHOUT REPETITIONS. THAT TOTAL NO OF WAYS IT CAN BE DONE IS?