Question

The number of ordered triplets $(a,b,c)$ where $a,b\in W$ and $c\in N$ such that origin and the point $(1,1,4)$ lie on the same side of the plane $ax+by+cz=39$ is

• A. $2280$
• B. $2190$
• C. $2180$
• D. $2290$

Answer 1

The correct option is B $2190$

Given plane is $P:ax+by+cz=39$
$\Rightarrow ax+by+cz-39=0$
Putting the coordinates of origin, we get
$P_{(0,0,0)}=-39$
Putting $(1,1,4)$, we get
$a+b+4c-39<0$
$\Rightarrow a+b+4c+t=38,$ where $t\in W$
When $c=1$
$a+b+t=34$
Number of ways of choosing $a$ and $b$ $={}^{34+3-1}{C}_{3-1}={}^{36}{C}_2$
When $c=2$
$a+b+t=30$
Number of ways of choosing $a$ and $b$ $={}^{32}{C}_2$
When $c=3$
$a+b+t=26$
Number of ways of choosing $a$ and $b$ $={}^{28}{C}_2$
.
.
.
When $c=9$
$a+b+t=2$
Number of ways of choosing $a$ and $b$ $={}^4{C}_2$
Total number of ordered triplets
$={}^{36}{C}_2+{}^{32}{C}_2+\cdots +{}^4{C}_2=\sum _{r=1}^9{}^{4r}{C}_2=\sum _{r=1}^9\frac{4r(4r-1)}{2}=8\sum _{r=1}^9{r}^2-2\sum _{r=1}^{9}r=8\times \frac{9\left(10\right)\left(19\right)}{6}-2\times \frac{9\left(10\right)}{2}=120\times 19-90=2190$