The digits $1$, $2$, $3$, $4$, $5$ and $6$ are written down in some order to form a six digit number. Then (a) how many such six digits number are even? and (b) how many such six digits number are divisible by $12$?
My attempt: "In some order" means that the six digits number we get do not have repeated digits. Then there are $6!$ possible digits, that is we can make $720$ six digits number. Then the number of even numbers must be end with one of $2$ or $4$ or $6$. Then the number of even numbers can be formed by the given digits are $3\times 5!=360$. Then it is almost done. Now for (b) we have to find the possible numbers of six digits which is divisible by $12$, that is the number must be divisible by both $3$ and $4$. I know that a number is divisible by $3$ if the sum of the numbers are divisible by $3$ and a number is divisible by $4$ if its last two digit is divisible by $4$. But how can I find that how many of such common numbers are there that are divisible by both three and four? Please help me to solve this.
Answer
Since digits are not repeated, all $6$ digits must be used.
Sum of all digits is $21$ $(1+2+3+4+5+6)$. So this number is divisible by $3$.
Now we only have to check its divisibility by $4$.
Number should end with any of $8$ combination $(12,16,24,32,36,52,56,64)$.
Starting $4$ digits can be in any order.
So total count $= 8*4! = 192$
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