In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where
the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3;
2+2+2+2+2+2;2+4+6; and many more
 A: You are looking for number of partitions of 12 in parts greater than 1. Because their number is not so big we can present all of them in a list bellow
$$\begin{array}{c|c}
2+2+2+2+2+2 & 1 \\
2+2+2+2+4 &  5 \\
2+2+2+3+3 &  10 \\
2+2+2+6 &  4 \\
2+2+3+5 & 12 \\
2+2+4+4 & 6 \\
2+3+3+4 & 12 \\
3+3+3+3 & 1 \\
2+2+8 & 3 \\
2+3+7 & 6 \\
2+4+6 & 6 \\
2+5+5 & 3 \\
3+3+6 & 3 \\
3+4+5 & 6 \\
4+4+4 & 1 \\
2+10 & 2 \\
3+9 & 2 \\
4+8 & 2 \\
5+7 & 2 \\
6+6 & 1 \\
12 & 1 \\
\end{array}$$
The partition turn in composition if we rearrange them. Number of arrangements is given on the right, so total number of compositions is $89$
A: Following answer is for general case.
Denote by $c_m(1,n)=\binom{n-1}{m-1}$ the number of composition of $n$ into $m$ positive parts, and by $c_m(2,n)$ number of compositions of $n$ into $m$ parts greater than $1$. 
Each composition of $n$ into $m$ parts greater than $1$, if all parts decrease by 1, becomes a composition of $n-m$ into $m$ positive parts. That means $$c_m(2,n)=c_m(1,n-m)=\binom{n-m-1}{m-1}$$
Number of all compositions of $n$ into parts greater than $1$ is $$c(2,n)=\sum_{m=1}^{n}c_m(2,n)=\sum_{m=1}^{n}\binom{n-m-1}{m-1}$$
In our case $n=12$
$$c(2,12)=\sum_{m=1}^{12}\binom{11-m}{m-1}=\sum_{m=1}^{6}\binom{11-m}{m-1}=$$
$$=\binom{10}{0}+\binom{9}{1}+\binom{8}{2}+\binom{7}{3}+\binom{6}{4}+\binom{5}{5}=$$
$$1+9+28+35+15+1=89$$ 
A: Let's split the question into a few pieces.
First part: how many ways to sum if all numbers are even? In this case, it's a simple partition problem, and the answer is equally simple. If you imagine the digits like this:
..|..|..|..|..|.. = 2+2+2+2+2+2
Then each of the five dividers can either be there or not, giving exactly $2^5=32$ ways of summing when they're all even.
Second part: Having the odd part summing to six, how many ways? In this case, there's just 3+3, and then you place three 2s and two 3s, which is equivalent to choosing two places in a sequence of five to place the 3s. But you can also place a 2 and a 4, a 4 and a 2, or a 6. So that's $\binom{5}{2}+2\binom{4}{2}+\binom{3}{2}=10+2\cdot6+3=25$. So far, we have 57.
Third part: Having the odd part summing to eight, how many ways? In this case, there's 3+5 and 5+3, and either two 2s or a 4. So that's $2\binom{4}{2}+2\binom{3}{2}=2\cdot 6+2\cdot 3 = 18$. We're up to 75.
Fourth part: Having the odd part summing to ten, how many ways? In this case, there's 3+7, 5+5, and 7+3, and then a 2 in each case. So there's three ways of placing the 2, and then three ways of choosing the remaining number pair, for a total of 9. That brings us to 84.
Final part: Having the odd part summing to twelve, how many ways? In this case, it's easiest to count - there's 3+3+3+3, 3+9, 5+7, 7+5, and 9+3, for a total of 5. So our answer is 89.
A: All the posted answers are answering the following variation of the original question:
"In how many ways i can write 12 as an ordered sum of integers where the smallest of those integers is at least 2?". But, the question asks the count when the smallest of the integers is equal to 2, so all the posted answers are over-counting (polite way to say that they are incorrect) -- for example, 3+4+5 or 6+6 are not valid compositions since the smallest integer is 3 and 6, respectively, in these examples, not 2. Discounting such compositions from Adi Dani's enumerative solution above, I get 70.  
