The correct option is C 105
x+y+z=16
When x is even.
Then there are two cases.
Case 1: x,y,z are even.
Case 2: x is even and y,z both are odd.
Case 1:
The equation reduces to
2x′+2y′+2z′=16, x′,y′,z′>0
⇒x′+y′+z′=8
Number of solutions = 7C2=21
Case 2:
The equation reduces to
2x′+2y′−1+2z′−1=16, x′,y′,z′>0
⇒x′+y′+z′=9
Number of solutions = 8C2=28
∴m=21+28=49
When y is odd.
This case is similar to Case 2 with either x or y being even or odd respectively.
∴ Number of solutions, n=2×8C2=56
So, m+n=105
Alternate Solution––––––––––––––––––––
x+y+z=16
Considering x is even.
If x=2, then y+z=14
Number of Solutions=13C1=13
If x=4, then y+z=12
Number of Solutions=11C1=11
If x=6, then y+z=10
Number of Solutions=9C1=9
Proceeding similarly, number of solutions for x=8,10,12,14 are 7,5,3,1
m=13+11+9+7+5+3+1=49
Now, considering y is odd.
If y=1, then x+z=15
Number of Solutions=14C1=14
If y=3, then x+z=13
Number of Solutions=12C1=12
If y=5, then x+z=11
Number of Solutions=11C1=10
Proceeding similarly, number of solutions for y=7,9,11,13 are 8,6,4,2
n=14+12+10+8+6+4+2=56
m+n=105