Question

If the number of solutions of the equation $x+y+z=20,$ where $1\leq x < y < z$ and $x,y,z\in I$ is $k$ , then $\frac{k}{10}$ is equal to

Answer 1

Let, $y=x+h,$ where $h\geq 1$ and $z=x+h+k,$ where $k\geq 1.$
Then, $x+y+z=20$
$\Rightarrow x+x+h+x+h+k=20$
$\Rightarrow 3x+2h+k=20.$
When $x=1\Rightarrow 2h+k=17,$ then there are $8$ ways.
When $x=2\Rightarrow 2h+k=14,$ then there are $6$ ways.
When $x=3\Rightarrow 2h+k=11,$ then there are $5$ ways.
When $x=4\Rightarrow 2h+k=8,$ then there are $3$ ways.
When $x=5\Rightarrow 2h+k=5,$ then there are $2$ ways.
When $x>5\Rightarrow $ not possible.
$\Rightarrow $ Total number of ways $=24$