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Question Number 116881 by bemath last updated on 07/Oct/20
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{m}\:\mathrm{of}\:\mathrm{non}\:\mathrm{negative} \\ $$$$\mathrm{integer}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{18} \\ $$$$ \\ $$
Answered by john santu last updated on 07/Oct/20
$${we}\:{can}\:{view}\:{each}\:{solution}\:,{say}\:{x}=\mathrm{3},\:{y}=\mathrm{7}\:,{z}=\mathrm{8} \\ $$$${as}\:{a}\:{combination}\:{of}\:{r}\:=\:\mathrm{18}\:{objects} \\ $$$${consisting}\:{of}\:\mathrm{3}\:{a}'{s}\:,\:\mathrm{7}\:{b}'{s}\:{and}\:\mathrm{8}\:{c}'{s},\:{where}\: \\ $$$${there}\:{are}\:{M}\:=\:\mathrm{3}\:{kinds}\:{of}\:{objects}\:{a}'{s},\:{b}'{s}\: \\ $$$${and}\:{c}'{s}\:.\:{Then}\:{m}\:=\:{C}\left({r}+{M}−\mathrm{1},{M}−\mathrm{1}\right) \\ $$$$\:=\:{C}\left(\mathrm{20},\mathrm{2}\right)\:=\:\frac{\mathrm{20}×\mathrm{19}}{\mathrm{2}×\mathrm{1}}\:=\:\mathrm{190} \\ $$
Answered by mr W last updated on 07/Oct/20
$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${k}=\mathrm{18}: \\ $$$${C}_{\mathrm{2}} ^{\mathrm{20}} =\frac{\mathrm{20}×\mathrm{19}}{\mathrm{2}}=\mathrm{190} \\ $$
Answered by mr W last updated on 07/Oct/20
$${x}+{y}={n} \\ $$$${x}=\mathrm{0},\mathrm{1},...,{n}\:\Rightarrow{n}+\mathrm{1}\:{solutions} \\ $$$$ \\ $$$${n}=\mathrm{18}−{z}=\mathrm{0}..\mathrm{18} \\ $$$$ \\ $$$$\Rightarrow{totally}\:\mathrm{1}+\mathrm{2}+...+\mathrm{19}=\frac{\mathrm{19}×\mathrm{20}}{\mathrm{2}}=\mathrm{190} \\ $$