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

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(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 +

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$^mC_{k-1}$

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(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

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