Question

Let $x+y+z+w=30;$ where $x,y,z,w\in N.$ If $m$ and $n$ denote the number of solutions when no variable may exceed $10$ and each variable is an odd number respectively, then $m+n=$

Answer 1

When no variable may exceed $10$
The number of solutions = coefficient of $x^{30}$ in
$(x+x^2+x^3+...+x^{10}{)}^4$
= coeff. of $x^{30}$ in ${\left[\frac{x(1-x^{10})}{1-x}\right]}^4$
= coeff. of $x^{30}$ in $x^4(1-x^{10}{)}^4(1-x{)}^{-4}$
= coeff. of $x^{30}$ in $x^4(1-4x^{10}+6x^{20}-...)(1-x{)}^{-4}$
= coeff. of $x^{30}$ in $(x^4-4x^{14}+6x^{24})(1-x{)}^{-4}$

Now, coeff. of $x^{26}$ in $(1-x{)}^{-4}={}^{29}{C}_3$
coeff. of $x^{16}$ in $(1-x{)}^{-4}={}^{19}{C}_3$
coeff. of $x^6$ in $(1-x{)}^{-4}={}^9{C}_3$

The number of required solutions is,
$m={}^{29}{C}_3-4\cdot {}^{19}{C}_3+6\cdot {}^9{C}_3=282$

When each variable is an odd number
put $x=2n+1,y=2p+1,z=2q+1,w=2r+1$
$x,y,z,w\ge 1\&x+y+z+w=30$
$\Rightarrow (2n+1)+(2p+1)+(2q+1)+(2r+1)=30$
$\Rightarrow n+p+q+r=13\&n,p,q,r\ge 0$
The number of required solutions is,
$={}^{13+4-1}{C}_{4-1}={}^{16}{C}_3=560$

$\therefore m+n=842$