Question: **The Question**

![enter image description here](https://i.stack.imgur.com/24Brr.png)

**My Problem**

Part a is straight forward, just $C(35,32)$. I'm having a little difficulty with the restrictions and understanding what they mean. $x\_1 > 0$ means we shouldn't have any solutions of the form $(0 + 32)$? If that's the case how would I factor that into my answer? I'm thinking taking the answer to part a and subtracting all the solutions which contain $32+0$ but I don't really know how to count those.

Answer: 
For part (b), $x\_i > 0$ for all $i$, let $y\_i = x\_i - 1$. Then the problem is equivalent to problem (a), only with a sum of 28 instead of 32 -- the answer is $C(31, 28) == C(31,3)$.

For part (c) let $y\_i = x\_i-5$ for $i\in \left\{ 1,2 \right\} $ and $y\_i = x\_i-7 $ otherwise. Then you have problem (a) again, with a sum of 8; the answer is $C(7,3)$.

For part (d) the same reasoning gives $C(3,3) = 1$.

For part (e) let $y\_i = x\_i +2$ to get $C(39,3)$.

For part (f), you can start with the solution to (b), and subtract the cases where $x\_4 > 25$. To get the latter, do $y\_4 = x\_4 -25$, and find an answer of $C(10,3)$ so the answer to part (f) is $C(35,3) - C(10,3)$.
For part f) we can use generating functions. The one that models f is

$$G(x)=\left(x+x^2+x^3+...+\right)^3 \left(x+x^2+...+x^{25}\right)$$

these are useful because they reduce combinatoric problems to computational ones. The above gf can be written:

$$g=\left(\frac{x}{1-x}\right)^3 \cdot \frac{ \left(x-x^{26}\right)}{1-x}$$

Now we just need to get at the coefficient of $x^{32}$. I have done that already for you to keep the post small. The term of interest is $4475x^{32}$. So the answer is 4475.