<s>[INST] 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
 [/INST] 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.
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