JEE Main 2023 — Mathematics Algebra

Given,

$$x+y+z=15$$x+y+z=15

Now we know that,

Non-negative integral solution of equation $a+b+c=n$a+b+c=n is given by $C3-1n+3-1$C3−1n+3−1 where a=b=c&a=b\neq c are also possibilities

So by above formula we get,

Total number of non-negative solution will be, $C3-115+3-1=C217$C3−115+3−1=C217

Now solving If any of these $2$2 are equal

So, the equation will become$x+2y=15$x+2y=15

Now finding possible cases we get,

$$\begin{matrix}y=0 & x=15\end{matrix}$$y=0​x=15​

$$\begin{matrix}y=1 & x=13\end{matrix}$$y=1​x=13​

$$\begin{matrix}y=2 & x=11\end{matrix}$$y=2​x=11​

$$\begin{matrix}y=3 & x=9\end{matrix}$$y=3​x=9​

$$\begin{matrix}y=4 & x=7\end{matrix}$$y=4​x=7​

$$y=5x=5\rightarrow x=y=z=5$$y=5x=5→x=y=z=5

$$y=6x=3$$y=6x=3

$$y=7x=1$$y=7x=1

So, total possibilities where x,y&z are distinct will be,

$$=C217-C23\times 8+2$$=C217−C23×8+2

{Note adding $2$2 because the cases $x=y=z$x=y=z is subtracted three times}

$$=136-24+2=114$$=136−24+2=114