Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Let n and k be positive integers such that $n \geq \frac{k(k+1)}{2}$ . The number of solutions $\left(x_1, x_2, \ldots, x_k\right)$ , all integers $x_1 \geq 1, x_2 \geq 2, \ldots x_k \geq k$ , satisfying $x_1+x_2+\ldots+x_k=n$ , is (Find in terms of m=k + n - r - 1)
Option: 1
${ }^m C_{k-1}$

Option: 2
${ }^m C_{k+1}$

Option: 3
${ }^m C_{k}$

Option: 4
None of these

Answers (1)

best_answer

The number of solutions of $x_1+x_2+\ldots+x_k=n$

= Coefficient of $t^n$ in $\begin{array}{r} \left(t+t^2+t^3+\ldots\right)\left(t^2+t^3+\ldots\right) \dots \left(t^k+t^{k+1}+\ldots\right) \end{array}$

= Coefficient of $t^n$ in $t^{1+2+\ldots+k}\left(1+t+t^2+\ldots\right)^k$

But $1+2+\ldots+k=\frac{1}{2} k(k+1)=r$ (say)

and $1+t+t^2+\ldots=\frac{1}{(1-t)}$

Thus number of required solutions

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

= Coefficient of $t^{n-r}$ in

$\begin{aligned} & {\left[1+{ }^k C_1 t+{ }^{k+1} C_2 t^2+{ }^{k+2} C_3 t^3+\ldots\right]} \\ & ={ }^{k+n-r-1} C_{k-1}={ }^{k+n-r-1} C_{k-1}={ }^m C_{k-1}, \end{aligned}$ (where m=k + n - r - 1)

$\begin{aligned} & =k+n-1-\frac{1}{2} k(k+1)=\frac{1}{2}\left[2 k+2 n-2-k^2-k\right] \\ & =\frac{1}{2}\left(2 n-k^2+k-2\right) \end{aligned}$

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question