No. of positive integral solutions for $w+x+y+z=20$ under the following conditions 
No. of positive integral solutions for $w+x+y+z=20$ under the following conditions:
  (a)w,x,y,z are whole numbers
  (b)w,x,y,z are natural numbers
  (c)w,x,y,z are odd natural numbers
  (b)w,x,y,z are part of ${1,2,3,\cdots,10}$     

I was able to do the first and the second part but got stuck on the third and fourth. I am able to calculate the no. of solutions for w,x,y,z even but not odd. But subtracting from the universal set won't give the right answer.
I can't think of any way to begin the fourth one. It would be great if I could get a hint to solve these two questions.
 A: 
How many solutions does the equation $w + x + y + z = 20$ have in the positive odd integers?

As @SakethMalyala suggested in the comments, let 
\begin{align*}
w & = 2w' + 1\\
x & = 2x' + 1\\
y & = 2y' + 1\\
z & = 2z' + 1
\end{align*}
Then $w'$, $x'$, $y'$, and $z'$ are nonnegative integers.  Substituting for $w$, $x$, $y$, and $z$ in the equation 
$$w + x + y + z = 20 \tag{1}$$
yields
\begin{align*}
2w' + 1 + 2x' + 1 + 2y' + 1 + 2z' + 1 & = 20\\
2w' + 2x' + 2y' + 2z' & = 16\\
w' + x' + y' + z' & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers, which you evidently know how to solve.

 $$\binom{8 + 3}{3} = \binom{11}{3}$$

How many solutions does the equation $w + x + y + z = 20$ have in the positive integers if $w, x, y, z \leq 10$?
Solve the problem in the positive integers.  From these, exclude those solutions in which one or more of the variables exceeds $10$.  Notice that at most one of the variables could exceed $10$ since $2 \cdot 11 = 22 > 20$.  
Suppose $w > 10$.  Since $w$ is a positive integer, $w \geq 11$.  Hence, $w' = w - 10$ is a positive integer.  Moreover,
\begin{align*}
w + x + y + z & = 20\\
w' + 10 + x + y + z & = 20\\
w' + x + y + z & = 10 \tag{3}
\end{align*}
Equation 3 is an equation in the positive integers, which you evidently know how to solve.  By symmetry, there are an equal number of solutions for which $x$, $y$, or $z$ exceeds $10$.  
Hence, the number of solutions of equation 1 in the positive integers in which none of the variables exceeds $10$ is the number of solutions of equation 1 in the positive integers minus four times the number of solutions of equation 3 in the positive integers. 

 $$\binom{19}{3} - \binom{4}{1}\binom{9}{3}$$

A: Here's what I came up with for part d). A brute force method is to first look at pairs, then triplets, then quaduplets adding to 20 using only the numbers 1 to 10, never zero. 
For pairs $x+y=20$, the sums range from 2 to 20 and the number of ways to achieve sum $n$ is equal to $n-1$ up to $n=11$, (i.e. there are 10 ways to get a sum of 11, 1+10....to 10+1), then back down to 1 way to make a sum of 20. The number of ways to sum to 12 will be the same as the number of ways to sum to 10, 13 to 9, etc. 
On to triplets now, $x+y+z=20$. There's similarly one way to make a sum of 3 and one way to make a sum of 30, 3 ways to make a sum of 4 and 3 ways to make a sum of 29. The maximum number of ways to achieve the sum will be for a sum of 17, which will have $(17-1)*(16-1)/2$ = 120 ways. Remember that between 3 and 30, two numbers that sum to 30 will have the same number of combinations of three numbers to achieve that sum, so 16 and 18 will have the same number, namely $(16-1)*(16-2)/2$, etc.
At this point you could find a formula (most likely a cubic form) for quadruplets, or just sum up the numbers you get from 3 to 19 as that is what you will add to the fourth number as it ranges from its maximum of 17 down to 1. i.e when $w$ is 3, there are 120 ways that $x$, $y$ and $z$ add up to 17. The total number of ways to add the four numbers I get is 680.
