Question

Consider the equation $x+y+z=16$ where $x,y,z>0$. Let $m$ and $n$ denote the number of positive integral solutions of the equation when $x$ is even and $y$ is odd, respectively. Then the value of $m+n$ is

• A. $49$
• B. $77$
• C. $105$
• D. $93$

Answer 1

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
$2x^{\prime }+2y^{\prime }+2z^{\prime }=16,x^{\prime },y^{\prime },z^{\prime }>0$
$\Rightarrow x^{\prime }+y^{\prime }+z^{\prime }=8$
Number of solutions $={}^7{C}_2=21$

Case $2:$
The equation reduces to
$2x^{\prime }+2y^{\prime }-1+2z^{\prime }-1=16,x^{\prime },y^{\prime },z^{\prime }>0$
$\Rightarrow x^{\prime }+y^{\prime }+z^{\prime }=9$
Number of solutions $={}^8{C}_2=28$

$\therefore m=21+28=49$

When $y$ is odd.
This case is similar to Case $2$ with either $x$ or $y$ being even or odd respectively.
$\therefore$ Number of solutions, $n=2\times {}^8{C}_2=56$

So, $m+n=105$

$\underset{–}{\text{Alternate Solution}}$
$x+y+z=16$
Considering $x$ is even.
If $x=2$, then $y+z=14$
Number of Solutions$={}^{13}{C}_1=13$
If $x=4$, then $y+z=12$
Number of Solutions$={}^{11}{C}_1=11$
If $x=6$, then $y+z=10$
Number of Solutions$={}^9{C}_1=9$
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$={}^{14}{C}_1=14$
If $y=3$, then $x+z=13$
Number of Solutions$={}^{12}{C}_1=12$
If $y=5$, then $x+z=11$
Number of Solutions$={}^{11}{C}_1=10$
Proceeding similarly, number of solutions for $y=7,9,11,13$ are $8,6,4,2$
$n=14+12+10+8+6+4+2=56$

$m+n=105$