How many 7-digit even numbers less than 3000000 can be formed using all the digits 1,2,2,3,5,5,6? I've somewhat got this question down but I'm only half way. 


  How many $7$-digit even numbers less than $3,000,000$ can be formed using all the digits $1,2,2,3,5,5,6$?


So I figured that there's about $4$ possible approaches

$1$ _ _ _ _ _ $6$
$1$ _ _ _ _ _ $2$
$2$ _ _ _ _ _ $6$
$2$ _ _ _ _ _ $2$  

How do I fill in the middle? I tried $5!$ and dividing out the similar factorials but I didn't get the right answer .
It appears that both previous answers missed that the number was to be even.  So the real count is 5!/2! + 5!/2!/2! + 5!/2! + 5!/2! = 210
Lets think that first digit is fixed due to which we get x __ __ __ __ __ __

Now these 6 places can be filled with 6 digits therefore total number of possibilities for the other 6 digits can be given as 6!

but here X has 2 possibilities so total no of possibilities = $$2*6!  = 1440$$

At last we have two 2's and two 5's so we get $$\frac{1440}{2!*2!} = 360$$