$6$ digit numbers formed from the first six positive integers such that they are divisible by both $4$ and $3$. 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.

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 $= 8*4! = 192$
