<s>[INST] 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.
 [/INST] 
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. 

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

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