Question

Let $n$ and $k$ be positive integers such that $n\ge \frac{k(k+1)}{2}.$ The number of solution $(x_1,x_2,..,x_k)\ge 1;x_2\ge 2,...,x_k\ge k$ all integers satisfying $x_1+x_2+x_3+...+x_k=n$ is

• A. ${}^m{C}_{k-1}$
• B. ${}^m{C}_k$3
• C. ${}^m{C}_{k+1}$
• D. None of these

Answer 1

Answer: (A)

${}^m{C}_{k-1}$

The number of solutions of $x_1+x_2+x_3+...x_k=n$

$=$ coefficients of $t^n$ in $(t+t^2+t^3+..)(t^2+t^3+..)...(t^k+t^{k+1}+..)$

$=$ coefficient of $t^n$ in $t^{1+2+3+...k}{(1+t+t^2+t^3+..)}^k$

But $1+2+3+...k=k\frac{k+1}{2}=r$

and $1+t+t^2+t^3+..=\frac{1}{1-t}$

Thus the required number of solutions

$=$ coefficients of $t^{n-r}$ in ${(1-t)}^{-k}$

$=$ coefficients of $t^{n-r}$ in$1{+}^k{C_{1}t+}^{k+1}{C}_2{t}^2+....$

$\Rightarrow {}^{k+n-r-1}{C}_{n-r}{=}^{k+n-r-1}{C}_{k-1}{=}^m{C}_{k-1}$