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Question Number 115769 by bemath last updated on 28/Sep/20
$${There}\:{are}\:\mathrm{3}\:{teachers}\:{and}\:\mathrm{6}\:{students} \\ $$$${who}\:{will}\:{sit}\:{on}\:{the}\:\mathrm{9}\:{available}\:{seats}.\:{many} \\ $$$${arrangements}\:{they}\:{sit}\:{if}\:{each}\: \\ $$$${teacher}\:{is}\:{flanked}\:{by}\:\mathrm{2}\:{students} \\ $$
Answered by mr W last updated on 28/Sep/20
$${method}\:\mathrm{1}: \\ $$$${S}\blacklozenge{S}\blacklozenge{S}\blacklozenge{S}\blacklozenge{S}\blacklozenge{S} \\ $$$${the}\:{three}\:{teachers}\:{may}\:{take}\:\mathrm{3}\:{of}\:{the} \\ $$$${places}\:{marked}\:{with}\:\blacklozenge.\:{there}\:{are} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{5}} \:{ways}. \\ $$$$\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{5}} ×\mathrm{3}!×\mathrm{6}!=\mathrm{43200} \\ $$$$ \\ $$$${method}\:\mathrm{2}: \\ $$$${S}_{\mathrm{1}} {TS}_{\mathrm{2}} {TS}_{\mathrm{3}} {TS}_{\mathrm{4}} \\ $$$${S}_{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}} \geqslant\mathrm{1} \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} +{S}_{\mathrm{4}} =\mathrm{6} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{4}} =\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }={x}^{\mathrm{4}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{6}} \:{is}\:{with}\:{k}=\mathrm{2} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{2}+\mathrm{3}} ={C}_{\mathrm{3}} ^{\mathrm{5}} \\ $$$$\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{5}} ×\mathrm{3}!×\mathrm{6}!=\mathrm{43200} \\ $$
Commented by bemath last updated on 29/Sep/20
$${gave}\:{kudos}.\:{santuy}\:{sir} \\ $$