Question

If $x+y+z+w=20$ where no variable may exceed $10,$ then the number of positive integral solutions is

Answer 1

When no variable may exceed $10,$
The number of solutions = coefficient of $x^{20}$ in $(x+x^2+x^3+\cdots +x^{10}{)}^4$
= coefficient of $x^{20}$ in ${\left[\frac{x(1-x^{10})}{1-x}\right]}^4$
= coefficient of $x^{20}$ in $x^4(1-x^{10}{)}^4(1-x{)}^{-4}$
= coefficient of $x^{20}$ in $x^4(1-4x^{10}+6x^{20}-\cdots )(1-x{)}^{-4}$

Now,
Coefficient of $x^{16}$ in $(1-x{)}^{-4}={}^{19}{C}_3$
Coefficient of $x^6$ in $(1-x{)}^{-4}={}^9{C}_3$
Number of required solutions
$={}^{19}{C}_3-4\cdot {}^9{C}_3=633$

Alternatively,
$x+y+z+w=20$
Total number of positive solutions $={}^{20-1}{C}_{4-1}={}^{19}{C}_3$
Let one variable be greater than $10,$ say $x.$
Put $x=x^{\prime }+10,$ where $x\in N$
Then, $x^{\prime }+y+z+w=10$
Number of positive solutions $={}^{10-1}{C}_{4-1}={}^9{C}_3$
Similarly, we can solve if other variables are greater than $10$
Hence, required number of solutions $={}^{19}{C}_3-4\cdot {}^9{C}_3$