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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$27$

Vedclass · Answer

(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) = $

Similar Questions

$A$ company has $10$ employees. The company has decided to form a team including at least three employees and also excluding at least three employees. Then the number of ways of forming the team is

Compute $\frac{12!}{10! \times 2!}$

If ${}^n C_7 = {}^n C_6$,then ${}^n C_2 = $

Compute $\frac{8!}{6! \times 2!}$

$A$ box contains $2$ white balls,$3$ black balls,and $4$ red balls. In how many ways can $3$ balls be drawn from the box if at least one black ball is to be included in the draw?

Vedclass Test Series

Exam Paper Generator

Online Exam Module