Question

Find the number of solutions of x + y+ z = 17, where x, y, z are non negative
integers such that x ϵ [2,5], y ϵ [3,6], z ϵ [4,7].

Answer 1

$\Large\textrm{Solution pairs are three.}$

• $(x,\ y,\ z)=(4,\ 6,\ 7)$

• $(x,\ y,\ z)=(5,\ 5,\ 7)$

• $(x,\ y,\ z)=(5,\ 6,\ 6)$

$\Large\textrm{We are given that: -}$

• $2\leq x\leq5$

• $3\leq y\leq 6$

• $4\leq z\leq7$

$\Large\textrm{To find: -}$

Number of solution pairs to $x+y+z=17$

$\Large\textrm{We know,}$

$x+y=17-z\ \dots\ (1)$

$5\leq x+y\leq 11$ and $10\leq x+y\leq13\ \because(1)$

$\therefore 10\leq x+y\leq 11$

$\Large\textrm{Possible cases}$

• $(x+y)+z=10+7$

• $(x+y)+z=11+6$

$\therefore z=6,\ 7$

$\Large\textrm{We know,}$

$y+z=17-x\ \dots\ (2)$

$7\leq y+z\leq13$ and $10\leq y+z\leq 13\ \because\ (2)$

$\therefore 10\leq y+z\leq 13$

$\Large\textrm{Possible cases}$

• $(y+z)+x=12+5$

• $(y+z)+x=13+4$

$\therefore x=4,\ 5$

$\Large\textrm{We know,}$

$z=x=17-y\ \dots\ (3)$

$6\leq z+x\leq 12$ and $11\leq z+x\leq 14\ \because\ (3)$

$\therefore 11\leq z+x\leq 12$

$\Large\textrm{Possible cases}$

• $(z+x)+y=11+6$

• $(z+x)+y=12+5$

$\therefore y=5,\ 6$

$\Large\textrm{Hence,}$

• $(x,\ y,\ z)=(4,\ 6,\ 7)$

• $(x,\ y,\ z)=(5,\ 5,\ 7)$

• $(x,\ y,\ z)=(5,\ 6,\ 6)$