When no variable may exceed 10
The number of solutions = coefficient of x30 in
(x+x2+x3+...+x10)4
= coeff. of x30 in [x(1−x10)1−x]4
= coeff. of x30 in x4(1−x10)4(1−x)−4
= coeff. of x30 in x4(1−4x10+6x20−...)(1−x)−4
= coeff. of x30 in (x4−4x14+6x24)(1−x)−4
Now, coeff. of x26 in (1−x)−4=29C3
coeff. of x16 in (1−x)−4=19C3
coeff. of x6 in (1−x)−4=9C3
The number of required solutions is,
m=29C3−4⋅19C3+6⋅9C3=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≥1 & x+y+z+w=30
⇒(2n+1)+(2p+1)+(2q+1)+(2r+1)=30
⇒n+p+q+r=13 & n,p,q,r≥0
The number of required solutions is,
= 13+4−1C4−1=16C3=560
∴m+n=842