Combinatorics question with unequality and different subscript 
a-) $x_1+x_2+...+x_7 \leq 30$ where $x_i's$ are even non-negative integers.
b-) $x_1+x_2+...+x_7 \leq 30$ where $x_i's$ are odd non-negative integers.
c-) $x_1+x_2+...+x_6 \leq 30$ where $x_i's$ are odd non-negative integers.

These questions is from my textbook. I know to solve similar questions such that $x_1+x_2+...+x_k \leq n$ where $x_i's$ are  non-negative integers. We add an extra term on the lefthandside and the rest found by combination with repetition such that $ \binom{n+(k+1)-1}{n}$.However , what happens when they are even or odd , is there any special technique ? Moreover , as you see the part a and b differ from only subscripts , i think that there must be a reason behind different subscript in same question.Is there any reason ? Thanks in advance..
 A: Finding the number of solutions to $x_1+x_2+\dots+x_7\leq 30$ where each of the $x_i$'s are even non-negative integers...
Rather than writing them as $x_1,x_2,\dots$ since they are even we can write them as $2y_1,2y_2,\dots$
We have then $2y_1+2y_2+\dots+2y_7\leq 30$.  Divide both sides by $2$ and we are left with $y_1+y_2+\dots+y_7\leq 15$ where each of the $y_i$'s are non-negative integers with no restrictions on parity.
Similarly, for the later problems where you deal with odd numbers, you can rewrite as $2z_1+1, 2z_2+1,\dots$  You can then subtract the $1$'s from both sides and then use the same idea.  The punchline is to rewrite a problem you haven't seen before in such a way that it becomes a problem you have seen before.
A: For $(a)$, you should solve for $y_1+y_2+...+y_7 \leq 15, ~ $ where $x_i = 2y_i, y_ \geq 0$
For $(b)$, you should solve for $y_1+y_2+...+y_7 \leq 11$ where $x_i = 2y_i + 1, y_i \geq 0$
For $(c)$, you should solve for $y_1+y_2+...+y_6 \leq 12$ where $x_i = 2y_i + 1, y_i \geq 0$
The difference between $(b)$ and $(c)$ is that in $(b)$, there are $7$ odd numbers and $7$ odd numbers cannot sum to an even number so equality $( = 30)$ is not possible and you would have sum $\leq 29$. But in $(c)$, you have $6$ odd numbers so equality can be reached.
