Let S_r = {(x, y, z) : x + y + z = 11, x geq | vedclass

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(B) Given,$S_r = \{(x, y, z) : x + y + z = 11, x \geq r, y \geq r, z \geq r\}$.For $S_2$: $x+y+z=11, x \geq 2, y \geq 2, z \geq 2$.Let $x-2=a, y-2=b,

Vedclass · Accepted Answer

$27$

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(B) Given,$S_r = \{(x, y, z) : x + y + z = 11, x \geq r, y \geq r, z \geq r\}$.For $S_2$: $x+y+z=11, x \geq 2, y \geq 2, z \geq 2$.Let $x-2=a, y-2=b, z-2=c$,where $a, b, c \geq 0$.Then $(a+2)+(b+2)+(c+2)=11 \Rightarrow a+b+c=5$.The number of non-negative integer solutions is given by $\binom{n+k-1}{k-1} = \binom{5+3-1}{3-1} = \binom{7}{2} = 21$.So,$n(S_2) = 21$.For $S_3$: $x+y+z=11, x \geq 3, y \geq 3, z \geq 3$.Let $x-3=a, y-3=b, z-3=c$,where $a, b, c \geq 0$.Then $(a+3)+(b+3)+(c+3)=11 \Rightarrow a+b+c=2$.The number of solutions is $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.So,$n(S_3) = 6$.For $S_4$: $x+y+z=11, x \geq 4, y \geq 4, z \geq 4$.Let $x-4=a, y-4=b, z-4=4$,where $a, b, c \geq 0$.Then $(a+4)+(b+4)+(c+4)=11 \Rightarrow a+b+c=-1$.Since $a, b, c \geq 0$,this is not possible,so $n(S_4) = 0$.Therefore,$n(S_2) + n(S_3) + n(S_4) = 21 + 6 + 0 = 27$.

Let $S_r = \{(x, y, z) : x + y + z = 11, x \geq r, y \geq r, z \geq r, x, y, z, r \in \mathbb{Z}\}$ and $n(S_r)$ represents the number of elements in $S_r$. Then $n(S_2) + n(S_3) + n(S_4) = $

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