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 .
 A: 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
A: We actually have only two cases: 1... and 2... .
In the first case, we have to fit the numbers 2,2,3,5,5,6 somehow. Imagine that we tag each of the repeated elements somehow, e.g. 2',2",3,5',5",6. If we care about the tags, then there are $6!$ possible ways to arrange the numbers. Now each untagged arrangement corresponds to $2!2!$ different tagged arrangements (consider the order in which the tags on the $2$s and $5$s appear). In total, this gives $6!/2!2!= 180$.
In the second case, we have only one repeated element $5$ so there are $6!/2! = 360$ possibilities, for a total of $540$.
In general, if we have $k$ unique elements repeated $t_1,\ldots,t_k$ times (respectively), then the answer is going to be $(t_1+\cdots+t_k)!/t_1!\cdots t_k!$, using the same reasoning. This is known as a multinomial coefficient since it appears in the multinomial theorem:
$$(x_1 + \cdots + x_k)^n = \sum_{t_1 + \cdots + t_k = n} n!/t_1!\cdots t_k! x_1^{t_1} \cdots x_k^{t_k}$$
where the sum is over all non-negative integers summing to $n$. The multinomial coefficient is sometimes denoted $\binom{n}{t_1\ldots t_k}$, although one of these numbers is really redundant. When $k=2$ we get the binomial coefficient, and we usually omit $t_2$.
A: Try the multinomial coefficient.
A: Everything you have done is right on the money. Your computation method (if I understand it correctly) should work. The answer to your last question (regarding what should be "in-between") is:
For $1.....6$: $5! / 2! / 2! = 120 / 4 = 30$
For $1.....2$: $5! / 2! = 120 / 2 = 60$
For $2.....6$: $5! / 2! = 120 / 2 = 60$
For $2.....2$: $5! / 2! = 120 / 2 = 60$
$60 + 60 + 60 + 30 = 210$
Therefore, the answer to the first question, regarding the right number, is 210. You can easily check for yourself that this is the right answer, with a small computer program that looks something like this:
Iterate through numbers $1000000$ to $3000000$, with increments of $2$.
For each iteration:   


*

*Go through all digits in the number and count the number of times
each digit is appearing. 

*If digits 1, 2, 3, 5 and 6 appear the
required number of times, increment the result variable with one.

*Repeat.
Once the iteration is complete, the result variable should
contain the number 210.

A: 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 $$\frac{1440}{2!*2!} = 360$$
