Question

The number of non integral solution of the equation, $x+y+3z=33$ is

• A. $120$
• B. $135$
• C. $210$
• D. $520$

Answer 1

Answer: (C)

$210$

By the given constants

$\to$x,y,z are non-negative integers

$\to$ x$\ge$0, y$\ge$0, z$\ge$0

$\to$ $x+y+3z=33$

$⟹$ y= 33-3z-x$⟶$(i)

12 possible values of $z$ are$0,1,2,3,4,5,6,7,8,9$,

$\bullet$ of $z=0$ from (i), $y=33-x$, where $x$ can have values from $0,1,...33=34$ possibilities

$\bullet$ of$z=1$ from (i) , $y=33-3-x=30-x$ .$x$ can have $31$ possibility from $(0,1,2,..30)$

$\bullet$ if$z=2$ from (i) $y=33-6-x=27-x$, $x$ can have $28$ possibilities $(0,1,2,...27)$ and so on.

$\bullet$ of$z=11$ from (i) $y=33-33-x=-x$. Only one possible combination with $x=y=0$

$\therefore$ Total possibilites = $34+31+28+...+1$

This is an $AP$ with $a=34$ , $d=-3$ and $n=12$

$\therefore$ Total possibilites = $34+31+28+...+1$=$\frac{12}{2}\ast${(2$\ast$34)+(12-1)$\ast$(-3)}

=$6(68-33)=210$