Question

The number of solutions (x, y, z) to the equation x - y - z = 25, where x, y, and z are positive integers such that $x\le 40,y\le 12$, and $z\le 12$ is

• A. 101
• B. 99
• C. 87
• D. 105

Answer 1

The correct option is B

99

$x-y-z=25$ where x, y, z are positive integers such that $x\le 40,y\le 12,z\le 12,y+z=x-25\dots \dots 1$
Now, for LHS to have feasible solutions under the given conditions, RHS = (x - 25) can take values from a minimum 2 to the maximum of 15.
i.e. $2\le (y+z)\le 15$
Now, let's consider $y+z\le 15$
$y+z+k=15\dots 2$, where k is any variable.
Putting $y=(y^{\prime }+1),z=(z^{\prime }+1)$ in equation 2), we get
$y^{\prime }+1+z^{\prime }+1+k=15y^{\prime }+z^{\prime }+k=13$
Number of solutions for the above equation =${}^{(13+3-1)}{C}_{(3-1)}{=}^{15}{C}_2=15\times \frac{14}{2}=105$
But the above number of solutions contains solutions with $y^{\prime }\ge 12,z^{\prime }\ge 12$. Those have to removed.
Number of invalid solutions = 2 + 4 = 6 ( 2 solutions for RHS = 13 and 4 solutions for RHS = 12)
Hence, final number of possible solutions = 105 - 6 = 99.