Question

Find number of negative integral solution of equation $x+y+z=-12$

Answer 1

$x+y+z=-12$ where $x,y,z$ are negative integers.

$\therefore 12=1+1+10;1+2+9;1+3+8;1+4+7;1+5+6$;here if $x=1,y+z$ can have $5$ different values but for example $y=1,z=10$.Also $z=1,y=10$ is also possible.

So, with $x=1$ we can have $10$ possibilities

Similarly if $y=1$,$10$ possibilities and if $z=1$,$10$ possibilities.

If we take $x=2$ or any even number we will get $y=5$ and $z=5$ so we can't reverse so for $x=$even we can have only $9$ possibilities.

Now,$y=$even can also have only $9$ possibilities and $z=$even can also have only $9$ possibilities.So, for even there are $27$ possibilities.

$x\to$odd$(1,3,5,7,9,11)$

we get $6$ odd cases

$\Rightarrow 6\times 30=180$possibilities ($\because$ each odd case has $30$ possibilities.

When $x\to$even$(2,4,6,8,10)$ we hve $5$ even cases.

$\Rightarrow 5\times 27=135$ possibolities.

Now if $x=12,y=z=0$ or if $y=12$ or $z=12$ there are no possibiliites because $0$ is not a negative integer.

Total positive integers set for $x+y+z=12$ is $180+135=315$

So, for $x+y+z=-12$ there will be $315$ negative integral solutions.