Let n and k be positive integers such that n | vedclass

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

Question

(A) The number of solutions to $x_1 + x_2 + ... + x_k = n$ with $x_i \ge i$ is the coefficient of $t^n$ in the product $(t^1 + t^2 + ...) (t^2 + t^3 +

Vedclass · Accepted Answer

$^mC_{k-1}$

Vedclass · Answer

(A) The number of solutions to $x_1 + x_2 + ... + x_k = n$ with $x_i \ge i$ is the coefficient of $t^n$ in the product $(t^1 + t^2 + ...) (t^2 + t^3 + ...) ... (t^k + t^{k+1} + ...)$.This is the coefficient of $t^n$ in $t^{1+2+...+k} (1 + t + t^2 + ...)^k$.Let $r = 1 + 2 + ... + k = \frac{k(k+1)}{2}$.The expression becomes the coefficient of $t^{n-r}$ in $(1-t)^{-k}$.Using the binomial expansion $(1-t)^{-k} = \sum_{j=0}^{\infty} \binom{k+j-1}{k-1} t^j$,the coefficient of $t^{n-r}$ is $\binom{k+(n-r)-1}{k-1}$.Let $m = k + n - r - 1 = k + n - \frac{k(k+1)}{2} - 1 = \frac{2k + 2n - k^2 - k - 2}{2} = \frac{2n - k^2 + k - 2}{2}$.Thus,the number of solutions is $\binom{m}{k-1}$.

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

Similar Questions

The value of $r$ for which $^{20}C_r ^{20}C_0 + ^{20}C_{r-1} ^{20}C_1 + ^{20}C_{r-2} ^{20}C_2 + ... + ^{20}C_0 ^{20}C_r$ is maximum is

$^{14}C_4 + \sum_{j=1}^{4} {^{18-j}C_3}$ is equal to

In how many ways can $6$ '$+$' signs and $4$ '$$' signs be arranged in a row such that no two '$$' signs occur together?

$A$ student has to answer a multiple-choice question having $5$ alternatives in which two or more than two alternatives are correct. Then the number of ways in which the student can answer that question is

The number of integral solutions to the equation $x+y+z=21$,where $x \geq 1, y \geq 3, z \geq 4$,is equal to $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module