Splitting 2 different objects across 3 people with additional properties Sorry for somewhat vague title, but I really couldn't explain it any further in the title alone..
Here goes the problem:
Joe, Bob and Smith need to split pencils and erasers.
In how many ways Joe, Bob and Smith can split 14 erasers and 10 pencils so that Joe gets at least 1 pencil and maximum of 5 erasers and both Bob and Smith gets at least 3 erasers and maximum of 4 pencils..
Now I am familiar with combinations in general, did a hefty amount of problems on my own but this one seems hard because it's multi-layered (at least in my eyes) and would really like to know what would be a good workflow to attack this kind of, multi-layered problems. :) Like, what's the best, and most user-friendly way to decompose them to multiple simple problems and them combine them together from bottom-up approach to get the result.. If any of this makes sense.. :)
EDIT: So I'd like to try to solve for erasers using approach by vadim123.
Looking at erasers, we have $J_E\le 5$, $B_E\ge 3$,  $S_E\ge 3$, and  $$J_E+B_E+S_E=14$$
We now set $B_E'=B_E-3$, to account for the condition on $B_E$,
$S_E'=S_E-3$, to account for the condition on $S_E$ and leave $J_E\le 5$ $$J_E+B_E'+S_E'=8 ~~~~(\star)$$ 
The original $(\star)$, ignoring the upper bounds, has $\left(\!{3\choose 8}\!\right)$ solutions.  
For $J_E\ge 6$, we set $J_E'=J_E-6$ and solve $J_E'+B_E'+S_E'=2$, which has $\left(\!{3\choose 2}\!\right)$ solutions. the final answer is
$$\left(\!{3\choose 8}\!\right)-\left(\!{3\choose 2}\!\right)=45-6=39$$
 A: Looking at pencils alone, we have $J_P\ge 1$, $B_P\le 4$,  $S_P\le 4$, and  $$J_P+B_P+S_P=10$$
We now set $J_P'=J_P-1$, to account for the condition on $J_P$, to get $B_P\le 4, S_P\le 4$, and $$J_P'+B_P+S_P=9 ~~~~(\star)$$
To handle the upper bounds, we need inclusion-exclusion.  We count solutions to $(\star)$ ignoring the upper bounds, then subtract "bad" solutions to $(\star)$ where $B_P\ge 5$, then subtract "bad" solutions to $(\star)$ where $S_P\ge 5$, then add "bad" solutions to $(\star)$ where both $B_P\ge 5$ and $S_P\ge 5$.
Fortunately, all of these are done using similar methods.  The original $(\star)$, ignoring the upper bounds, has $\left(\!{3\choose 9}\!\right)$ solutions, where this is a multiset.  For $B_P\ge 5$, we set $B_P'=B_P-5$ and solve $J_P'+B_P'+S_P=4$, which has $\left(\!{3\choose 4}\!\right)$ solutions.  If $S_P\ge 5$ we get the same value.  Lastly, there are zero solutions where both $S_P\ge 5$ and $B_P\ge 5$.  Hence the final answer is
$$\left(\!{3\choose 9}\!\right)-2\left(\!{3\choose 4}\!\right)=25$$
Now, perform a similar calculation for erasers, then combine the two answers by multiplication.
