Orbital angular momentum quantum numbers - subtracted? Reading Griffiths' Quantum Mechanics. 
We have the electronic confirmation of Carbon as 
$$(1s)^2 (2s)^2 (2p)^2$$
in the ground state.
He says

There are two electrons with orbital angular momentum quantum number $1$, so the total angular momentum quantum number could be $2, 1$ or $0$. 

How? The angular momentum quantum number $l$ is positive. There are two electrons with $l=1$, so I can only see that you can combine them to get $L=2$. 
Not sure how the other options come about - only can see $0 = 1-1$, but I didn't think you would take them away to get the total angular momentum
 A: With two electrons with angular momentum quantum numer $l=1$ there are three possibilities, you can heuristically think of them like this:
$L=0$:  The angular momentum vectors of the different electrons are anti-aligned. So the total angular momentum is $1-1=0$.
$L=1$: The angular momentum vector of one of the electrons is pointing along the z-axis, while the others doesn't. So the total angular momentum for this configuration is $1+0=1$
$L=2$: The angular momentum vectors are both aligned, giving $1+1=2$.
You can also look at the $m_l$ quantum number. You always have two electrons with $l=1$, but the different possible combinations of $m_{l_1}$ and $m_{l_2}$ give you different total angular momentum:
($m_{l_1}=1$ and $m_{l_2}=-1$) or ($m_{l_1}=-1$ and $m_{l_2}=1$) gives $L=0$,
($m_{l_1}=1$ and $m_{l_2}=0$) or ($m_{l_1}=0$ and $m_{l_2}=1$) or ($m_{l_1}=-1$ and $m_{l_2}=0$) or ($m_{l_1}=0$ and $m_{l_2}=-1$) gives $L=1$,
($m_{l_1}=1$ and $m_{l_2}=1$) or ($m_{l_1}=-1$ and $m_{l_2}=-1$) gives $L=2$.
A: Let's review the simpler case of adding two spin 1/2 particles before doing the calculation for two spin 1 particles. The discussion hinges on the ladder operators crucially so maybe review those if you're unfamiliar.
Spin 1/2
The single particle states are spanned by $\{|+\rangle , |-\rangle\}$ so the two particle state has basis:
$$ \{ |--\rangle ,|-+\rangle , |+-\rangle , |++\rangle \}$$
note that two of these obey an exchange symmetry but the middle two do not. We can fix this by rewriting the middle two states as combinations of symmetric/anti-symmetric states:
$$|-+\rangle = \frac{1}{2}(|-+\rangle +|+-\rangle )+\frac{1}{2}(|-+\rangle -|+-\rangle )$$
and likewise for $|+-\rangle$, thus our two spin system has an exchange symmetry adapted basis:
$$ \{ |--\rangle ,\frac{1}{\sqrt{2}}(|-+\rangle +|+-\rangle ) , \frac{1}{\sqrt{2}}(|-+\rangle -|+-\rangle ) , |++\rangle \}$$
We can now ask the question: what is the total spin in each of these states? For this you need to know the total spin operator $J^2=(\vec{J}_1+\vec{J}_2)^2=J_1^2+J_2^2+2\vec{J}_1\cdot\vec{J}_2$ The last term can be written in terms of components as $2\vec{J}_1\cdot\vec{J}_2 = 2J_1^z1J_2^z + 2(J_1^xJ_2^x + J_1^yJ_2^y)$ and the last term can again be rewritten in terms of ladder operators:
$$2(J_1^xJ_2^x + J_1^yJ_2^y) = J_1^+J_2^- + J_1^-J_2^+$$
which finally allows one to write
$$J^2 = J_1^2+J_2^2 + 2J_1^zJ_2^z + J_1^+J_2^- + J_1^-J_2^+$$
(this expression is not limited to the $J=1/2$ case).
You can then simply apply this to each state and recover that the three symmetric basis states we wrote down are all $J=1$ states and the anti-symmetric states are $J=0$. 
Spin 1
In this case there are still two particles but now three states each. We get an unsymmetrised basis with $3^2=9$ states. These states are not eigenstates of the total angular momentum $J^2$ which should behave well under particle exchange. When all is said and done, we will get eight symmetry adapted states which break up into a set of 5 symmetric states ($J=2$) a set of 3 anti-symmetric states($J=1$) and a singlet symmetric state ($J=0$). 
You can think of the $J=0$ states as having spins that are anti-aligned and so cancelling out, but be wary of the fact that these states are in fact (anti)-symmetrised combinations of the single particle states that may make sense to you, and the relative phases play a big role in determining the final angular momentum. However, the intuitive picture does correctly predict that adding two spins $j_1, j_2$ gives a total spin $|j_1-j_2|\leq J \leq |j_1+j_2|$.
