Question Number 122250 by mr W last updated on 15/Nov/20
Answered by mr W last updated on 15/Nov/20
$${using}\:{generating}\:{function}\:{method} \\ $$$${the}\:{number}\:{of}\:{ways}\:{is}\:{the}\:{coefficient} \\ $$$${of}\:{term}\:{x}^{\mathrm{10}} \:{in}\:{the}\:{expansion} \\ $$$$\left({C}_{\mathrm{1}} ^{\mathrm{15}} {x}+{C}_{\mathrm{2}} ^{\mathrm{15}} {x}^{\mathrm{2}} +…+{C}_{\mathrm{15}} ^{\mathrm{15}} {x}^{\mathrm{15}} \right)× \\ $$$$\left({C}_{\mathrm{1}} ^{\mathrm{12}} {x}+{C}_{\mathrm{2}} ^{\mathrm{12}} {x}^{\mathrm{2}} +…+{C}_{\mathrm{12}} ^{\mathrm{12}} {x}^{\mathrm{12}} \right)× \\ $$$$\left({C}_{\mathrm{1}} ^{\mathrm{10}} {x}+{C}_{\mathrm{2}} ^{\mathrm{10}} {x}^{\mathrm{2}} +…+{C}_{\mathrm{10}} ^{\mathrm{10}} {x}^{\mathrm{10}} \right)× \\ $$$$\left({C}_{\mathrm{1}} ^{\mathrm{5}} {x}+{C}_{\mathrm{2}} ^{\mathrm{5}} {x}^{\mathrm{2}} +…+{C}_{\mathrm{5}} ^{\mathrm{5}} {x}^{\mathrm{5}} \right) \\ $$$${which}\:{is}\:\mathrm{1}\:\mathrm{032}\:\mathrm{675}\:\mathrm{125}. \\ $$$$ \\ $$$${general}\:{form} \\ $$$${GF}=\underset{{i}=\mathrm{1}} {\overset{{m}} {\prod}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}_{{i}} } {\sum}}{C}_{{k}} ^{{n}_{{i}} } {x}^{{k}} \right) \\ $$
Commented by mr W last updated on 15/Nov/20
$${if}\:{we}\:{choose}\:{totally}\:\mathrm{4}\:{books},\:{the} \\ $$$${answer}\:{is}\:\mathrm{15}×\mathrm{12}×\mathrm{10}×\mathrm{5}=\mathrm{9}\:\mathrm{000}. \\ $$$$\left(={coef}.\:{of}\:{term}\:{x}^{\mathrm{4}} \right)\:\checkmark \\ $$$$ \\ $$$${if}\:{we}\:{choose}\:{totally}\:\mathrm{5}\:{books},\:{the} \\ $$$${answer}\:{is} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{15}} ×\mathrm{12}×\mathrm{10}×\mathrm{5}+{C}_{\mathrm{2}} ^{\mathrm{12}} ×\mathrm{15}×\mathrm{10}×\mathrm{5}+ \\ $$$${C}_{\mathrm{2}} ^{\mathrm{10}} ×\mathrm{15}×\mathrm{12}×\mathrm{5}+{C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{15}×\mathrm{12}×\mathrm{10} \\ $$$$=\mathrm{171}\:\mathrm{000}. \\ $$$$\left(={coef}.\:{of}\:{term}\:{x}^{\mathrm{5}} \right)\:\checkmark \\ $$
Commented by mr W last updated on 15/Nov/20
