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Let $S_r=\{x, y, z) / x+y+z=11, x \geq r, y \geq r$, $z \geq r, x, y, z, r$ are integers $\}$ and $n\left(S_r\right)$ represents the number of elements in $S_r$. Then $n\left(S_{2)}+n\left(S_3\right)+n\left(S_4\right)=\right.$
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Verified Answer
The correct answer is:
27
Given,
$\begin{aligned}
& S_r=\{(x, y, z): x+y+z=11, x \geq r, y \geq r, z \geq r\} \\
& S_2=\{(x, y, z): x+y+z=11, x \geq 2, x \geq 2, z \geq 2\} \\
& \because x \geq 2 \Rightarrow x-2 \geq 0, y-2 \geq 0, z-2 \geq 0 \\
& \text { Let } x-2=a, y-2=b, z-2=c \\
& \therefore x+y+z=11 \Rightarrow a+b+c=5, a, b, c \geq 0 \\
& \quad n\left(S_2\right)={ }^{5+3-1} C_{3-1}={ }^7 C_2=21
\end{aligned}
$Similary, for $S_3, a+b+c=2$
$n\left(S_3\right)={ }^{2+3-1} C_2={ }^4 C_2=6$
For $S_4, a+b+c=-1 \quad$ (not possible)
$\therefore \quad n\left(S_2\right)+n\left(S_3\right)+n\left(S_4\right)=21+6+0=27$
$\begin{aligned}
& S_r=\{(x, y, z): x+y+z=11, x \geq r, y \geq r, z \geq r\} \\
& S_2=\{(x, y, z): x+y+z=11, x \geq 2, x \geq 2, z \geq 2\} \\
& \because x \geq 2 \Rightarrow x-2 \geq 0, y-2 \geq 0, z-2 \geq 0 \\
& \text { Let } x-2=a, y-2=b, z-2=c \\
& \therefore x+y+z=11 \Rightarrow a+b+c=5, a, b, c \geq 0 \\
& \quad n\left(S_2\right)={ }^{5+3-1} C_{3-1}={ }^7 C_2=21
\end{aligned}
$Similary, for $S_3, a+b+c=2$
$n\left(S_3\right)={ }^{2+3-1} C_2={ }^4 C_2=6$
For $S_4, a+b+c=-1 \quad$ (not possible)
$\therefore \quad n\left(S_2\right)+n\left(S_3\right)+n\left(S_4\right)=21+6+0=27$
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