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The no. of points $(x,y,z)$ in space whose coordinates are negative integers and $x+y+z+12=0$ is

$\begin{array}{1 1} 55 \\ 72 \\ 110 \\ 60 \end{array}$

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Combination of 3 negative integers whose sum is -12 are:
$(-1,-1,-10),(-1,-2,-9),(-1,-3,-8),(-1,-4,-7),(-1,-5,-6)$
$(-2,-2,-8),(-2,-3,-7),(-2,-4,-6),(-2,-5,-5)$
$(-3,-3,-6),(-3,-4,-5)$
$(-4,-4,-4)$
No. of points with the combination $(-4,-4,-4) = 1$
No. of points with the combinations $ (-1,-1,-10),(-2,-2,-8),(-2,-5,-5),(-3,-3,-6)$
= 3 in each combination=$3\times 4=12$
Remaining 7 combinations have $3!=6$ points in each combination.
$=7\times 6=42$ points
$\therefore$ Total number of required points in space =
$1+12+42=55$
answered Aug 15, 2013 by rvidyagovindarajan_1
 

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