Balls between boxes equation I have a practice question I don't know how to answer past part (a)

Consider nonnegative integer solutions of the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 26$
  (a) How many different solutions are there? [3 marks]
  (b) How many solutions also satisfy: for every $i \in \{1, 2, 3\}, x_i > 0?$ [3 marks]
  (c) How many solutions also satisfy: for every $i \in \{1, 2, 3, 4, 5, 6\}, x_i$ is even? [3 marks] 

For (a) I have ${31}\choose{26}$ and I feel confident in that, but I am unsure of how to answer the next two parts as I can't find anything in my notes that helps me with it.
Thanks
edit: think i have solved the second two parts but could someone tell me if they look right and if my logic is good.
I am considering the 6 unknown integers as boxes and the final answer of 26 as 26 balls.
part(b).
if i consider it as such, could it be that i just need to say that I place one ball in each of the first 3 boxes, leaving me with 23 to be split between all 6 of the boxes so that the answer would be C(28, 23)?
part (c).
I would now glue the balls together so that there are 13 pairs (each considered as one distinct object) to split between the 6 boxes, leaving me with C(18, 13)?
Apologies for the balls in boxes analogy but this is how I was taught this initially and I am trying to relate this to that, so I can have a better grasp of this. Thank you to anyone who has read this :)

The task in part b and part c is to find a way to reduce the problem to part a. 
Say for part b, here's a hint. Let: 
$$x_i^{'} = x_i - 1, \quad \forall i$$
Now if we find the non-negative solutions to below problem: 
$$\sum_{i=1}^{6} x_{i}^{'} = 32$$
we get all positive solutions to the original problem asked.
Which is very similar to part a. 
Similarly for part c, replace
$$x_i = 2y_{i} \quad \forall i$$,
where $y_i \in \{0,1,2,..., 13\}$. can you finish now?
