Question

If $n\text{\&}k$ be positive integers such that $n\ge \frac{k(k+1)}{2}$. The number of integral solutions $(x_1,x_2,...,x_k),x_1\ge 1,x_2\ge 2,...,x_k\ge k,$ satisfying $x_1+x_2+...+x_k=n$, is

• A. ${}^m{C}_{k-1},\text{where}m=\left(\frac{1}{2}\right)(2n-k^2+k-2)$
• B. ${}^m{C}_k,\text{where}m=\left(\frac{1}{2}\right)(2n-k^2+k-2)$
• C. ${}^m{C}_{k-1},\text{where}m=\left(\frac{1}{2}\right)(2n-k^2+2k-2)$
• D. ${}^m{C}_k,\text{where}m=\left(\frac{1}{2}\right)(2n-k^2+2k-2)$

Answer 1

Answer: (A)

${}^m{C}_{k-1},\text{where}m=\left(\frac{1}{2}\right)(2n-k^2+k-2)$

Given, $x_1\ge 1,x_2\ge 2,x_3\ge 3,.........x_k\ge k$
let, $x_1=y_1+1,x_2=y_2+2,x_3=y_3+3,..............x_k=y_k+k$, where $y_1,y_2,y_3,......y_k$ are non-negative integers
but, $x_1+x_2+........x_k=n$
$\Rightarrow$$y_1+1+y_2+2+y_3+3+................y_k+k=n$
$\Rightarrow y_1+y_2+y_3+................y_k+\frac{k(k+1)}{2}=n$
$\Rightarrow y_1+y_2+y_3+................y_k=n-\frac{k(k+1)}{2}$
$\therefore$ number of solutions=${}^{(n-\frac{k(k+1)}{2}+k-1)}{C}_{k-1}$
let m=$n-\frac{k(k+1)}{2}+k-1$
$\Rightarrow m=\left(\frac{1}{2}\right)(2n-k^2+k-2)$
Hence, option A is correct.