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Question Number 54860 by Tinkutara last updated on 13/Feb/19

Commented by Tinkutara last updated on 14/Feb/19

$$Butansweris$$$$\frac{1}{2}(2n-k^2+k-2)$$

Answered by kaivan.ahmadi last updated on 14/Feb/19

$$\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{if}{\mathrm{x}}_{\mathrm{i}}⩾0\mathrm{then}{\mathrm{x}}_1+...+{\mathrm{x}}_{\mathrm{k}}=\mathrm{n}$$$$\mathrm{has}\left(\begin{array}{c}\mathrm{n}+\mathrm{k}-1\\ \mathrm{k}-1\end{array}\right)\mathrm{answer}$$$$\mathrm{now}\mathrm{if}{\mathrm{x}}_{\mathrm{i}}⩾\mathrm{i}\mathrm{then}\mathrm{we}\mathrm{omit}1+...+\mathrm{k}=\frac{\mathrm{k}(\mathrm{k}+1)}{2}\mathrm{of}\mathrm{answers}$$$$\mathrm{so}\mathrm{we}\mathrm{have}$$$${\mathrm{x}}_1+...+{\mathrm{x}}_{\mathrm{k}}=\mathrm{n}-\frac{\mathrm{k}(\mathrm{k}+1)}{2}$$$$\mathrm{and}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{answer}\mathrm{is}$$$$\left(\begin{array}{c}\mathrm{n}-\frac{\mathrm{k}(\mathrm{k}+1)}{2}+\mathrm{k}-1\\ \mathrm{k}-1\end{array}\right)=\left(\begin{array}{c}\mathrm{n}-\frac{{\mathrm{k}}^2+\mathrm{k}}{2}-1\\ \mathrm{k}-1\end{array}\right)$$

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