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Question Number 224623 by gregori last updated on 22/Sep/25 | ||
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$$\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by gregori last updated on 23/Sep/25 | ||
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$${no}\:{sir}.\:{i}\:{don}'{t}\:{get}\:{answer}\: \\ $$ | ||
Commented by mr W last updated on 22/Sep/25 | ||
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$$\mathrm{8}\:\mathrm{467}\:\mathrm{200}\:{arrangements}? \\ $$ | ||
Answered by mehdee7396 last updated on 22/Sep/25 | ||
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$$\mathrm{6}!×\begin{pmatrix}{\mathrm{7}}\\{\mathrm{4}}\end{pmatrix}×\mathrm{4}!=\frac{\mathrm{6}!\mathrm{7}!}{\mathrm{3}!} \\ $$ | ||
Answered by mr W last updated on 24/Sep/25 | ||
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$$\mathrm{6}\:{men}\:{and}\:\mathrm{1}\:{empty}\:{chair}\:=\mathrm{7}\:{men} \\ $$$$\mathrm{4}\:{women}\:{and}\:\mathrm{7}\:{men}\:{sit}\:{in}\:\mathrm{11}\:{chairs} \\ $$$${AWBWBWBWA} \\ $$$${A}={zero}\:{or}\:{more}\:{men}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +... \\ $$$${B}={one}\:{or}\:{more}\:{men}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +... \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{2}} \left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{3}} \\ $$$$={x}^{\mathrm{3}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{\mathrm{5}} \\ $$$$=\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} }={x}^{\mathrm{3}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{7}} \:{is}\:{C}_{\mathrm{4}} ^{\mathrm{8}} \\ $$$$\Rightarrow{answer}\:{is}\:{C}_{\mathrm{4}} ^{\mathrm{8}} ×\mathrm{4}!×\mathrm{7}!=\mathrm{8}\:\mathrm{467}\:\mathrm{200}\:\checkmark \\ $$ | ||
Commented by mr W last updated on 24/Sep/25 | ||
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$${Alternative}\:{method}: \\ $$$${arrange}\:{the}\:\mathrm{6}\:{men}\:{and}\:{the}\:{empty} \\ $$$${chair}\:\left(=\:\mathrm{7}\:{men}\right): \\ $$$$\_{M\_M\_M\_M\_M\_M\_M\_} \\ $$$${there}\:{are}\:\mathrm{7}!\:{ways}\:{to}\:{do}\:{this}. \\ $$$${now}\:{arrange}\:{the}\:\mathrm{4}\:{women}.\:{each}\:{of} \\ $$$${them}\:{can}\:{take}\:{one}\:{of}\:{the}\:\mathrm{8}\:{places}\:``\_''. \\ $$$${there}\:{are}\:{C}_{\mathrm{4}} ^{\mathrm{8}} ×\mathrm{4}!={P}_{\mathrm{4}} ^{\mathrm{8}} \:{ways}\:{to}\:{do}\:{this}. \\ $$$${so}\:{the}\:{answer}\:{is}\:\mathrm{7}!×{P}_{\mathrm{4}} ^{\mathrm{8}} =\mathrm{8}\:\mathrm{467}\:\mathrm{200}. \\ $$ | ||