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Question Number 179738 by SLVR last updated on 01/Nov/22 | ||
$${Number}\:{of}\:{distributions}\:{of} \\ $$$$\mathrm{12}\:{different}\:{things}\:{be}\:{taken}\:{to} \\ $$$$\mathrm{3}{different}\:{boxes}\:{so}\:{as} \\ $$$$\left.\mathrm{1}\right)\mathrm{5}\:{things}\:{in}\:\mathrm{1}{st}\:{box}\:{exactly} \\ $$$$\left.\mathrm{2}\right)\mathrm{5}\:{things}\:{in}\:{any}\:{one}\:{box}? \\ $$ | ||
Answered by SLVR last updated on 01/Nov/22 | ||
$${I}\:{am}\:{getting}\:\mathrm{12}_{{C}_{\mathrm{5}} } ×\mathrm{2}^{\mathrm{7}} \:{for} \\ $$$$\mathrm{1}{st}\:{and}\:{but}\:\mathrm{2}{nd} \\ $$$${question}.=>>{please} \\ $$ | ||
Answered by mr W last updated on 03/Nov/22 | ||
$$\left.\mathrm{1}\right) \\ $$$$\mathrm{5}\:{things}\:{for}\:{the}\:\mathrm{1}{st}\:{box}:\:{C}_{\mathrm{5}} ^{\mathrm{12}} \\ $$$$\mathrm{7}\:{things}\:{into}\:{two}\:{boxes}:\:\mathrm{2}^{\mathrm{7}} \\ $$$$\Rightarrow{C}_{\mathrm{5}} ^{\mathrm{12}} ×\mathrm{2}^{\mathrm{7}} =\mathrm{101376} \\ $$$$\left.\mathrm{2}\right) \\ $$$${not}\:{clear},\:\:{what}\:{you}\:{mean}. \\ $$$${at}\:{least}\:{or}\:{at}\:{most}\:{or}\:{exactly}\:\mathrm{5}\:{things} \\ $$$${in}\:{one}\:{of}\:{the}\:{three}\:{boxes}? \\ $$ | ||
Commented by SLVR last updated on 02/Nov/22 | ||
$${sir}..{so}\:{kind}\:{of}\:{you}.. \\ $$$${here}\:{in}\:\mathrm{2}{nd}\:{i}\:{need}\:\mathrm{5}\:{exactly} \\ $$$$\left.{but}\:{what}\:{if}\:\:\:\:{a}\right){if}\:{at}\:{least}\:{one} \\ $$$${in}\:\mathrm{2}{nd}\:{box}? \\ $$$$\left.{b}\right){if}\:\:{at}\:{most}\:\mathrm{7}\:{in}\:\mathrm{2}{nd}\:{box} \\ $$$${please}\:{sir}\:{Prof}.{Mr}.{W} \\ $$ | ||
Commented by mr W last updated on 03/Nov/22 | ||
$$\left.\mathrm{2}{a}\right) \\ $$$${say}\:{in}\:{one}\:{and}\:{only}\:{one}\:{box}\:{there}\:{are} \\ $$$${exactly}\:\mathrm{5}\:{balls}. \\ $$$${to}\:{divide}\:\mathrm{12}\:{balls}\:{into}\:{three}\:{boxes} \\ $$$${there}\:{are}\:{following}\:{possibilities}: \\ $$$$\mathrm{5}+\mathrm{0}+\mathrm{7}\:\checkmark\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{5}!\mathrm{7}!}×\mathrm{3}! \\ $$$$\mathrm{5}+\mathrm{1}+\mathrm{6}\:\checkmark\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{5}!\mathrm{6}!}×\mathrm{3}! \\ $$$$\mathrm{5}+\mathrm{2}+\mathrm{5}\:× \\ $$$$\mathrm{5}+\mathrm{3}+\mathrm{4}\:\checkmark\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{5}!\mathrm{3}!\mathrm{4}!}×\mathrm{3}! \\ $$$$\Rightarrow{answer}\:{is}\: \\ $$$$\:\:\:\:\frac{\mathrm{12}!\mathrm{3}!}{\mathrm{5}!}×\left(\frac{\mathrm{1}}{\mathrm{7}!}+\frac{\mathrm{1}}{\mathrm{6}!}+\frac{\mathrm{1}}{\mathrm{3}!\mathrm{4}!}\right)=\mathrm{204}\:\mathrm{336} \\ $$ | ||
Commented by mr W last updated on 03/Nov/22 | ||
$$\left.\mathrm{2}{b}\right) \\ $$$${say}\:{in}\:{one}\:{box}\:{at}\:{least}\:\mathrm{5}\:{balls}: \\ $$$$\mathrm{12}+\mathrm{0}+\mathrm{0}\Rightarrow\mathrm{1}×\frac{\mathrm{3}!}{\mathrm{2}!}=\mathrm{3} \\ $$$$\mathrm{11}+\mathrm{1}+\mathrm{0}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{11}!}×\mathrm{3}!=\mathrm{72} \\ $$$$\mathrm{10}+\mathrm{2}+\mathrm{0}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{10}!\mathrm{2}!}×\mathrm{3}!=\mathrm{396} \\ $$$$\mathrm{10}+\mathrm{1}+\mathrm{1}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{10}!\mathrm{2}!}×\mathrm{3}!=\mathrm{396} \\ $$$$\mathrm{9}+\mathrm{3}+\mathrm{0}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{9}!\mathrm{3}!}×\mathrm{3}!=\mathrm{1}\:\mathrm{320} \\ $$$$\mathrm{9}+\mathrm{2}+\mathrm{1}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{9}!\mathrm{2}!}×\mathrm{3}!=\mathrm{3}\:\mathrm{960} \\ $$$$\mathrm{8}+\mathrm{4}+\mathrm{0}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{8}!\mathrm{4}!}×\mathrm{3}!=\mathrm{2}\:\mathrm{970} \\ $$$$\mathrm{8}+\mathrm{3}+\mathrm{1}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{8}!\mathrm{3}!}×\mathrm{3}!=\mathrm{11}\:\mathrm{880} \\ $$$$\mathrm{8}+\mathrm{2}+\mathrm{2}\:\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{8}!\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{3}!=\mathrm{8}\:\mathrm{910} \\ $$$$\mathrm{7}+\mathrm{5}+\mathrm{0}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{7}!\mathrm{5}!}×\mathrm{3}!=\mathrm{4}\:\mathrm{752} \\ $$$$\mathrm{7}+\mathrm{4}+\mathrm{1}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{7}!\mathrm{4}!}×\mathrm{3}!=\mathrm{23}\:\mathrm{760} \\ $$$$\mathrm{7}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{7}!\mathrm{3}!\mathrm{2}!}×\mathrm{3}!=\mathrm{47}\:\mathrm{520} \\ $$$$\mathrm{6}+\mathrm{6}+\mathrm{0}\:\Rightarrow\frac{\mathrm{12}!}{\left(\mathrm{6}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{3}!=\mathrm{2}\:\mathrm{772} \\ $$$$\mathrm{6}+\mathrm{5}+\mathrm{1}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{6}!\mathrm{5}!}×\mathrm{3}!=\mathrm{33}\:\mathrm{264} \\ $$$$\mathrm{6}+\mathrm{4}+\mathrm{2}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{6}!\mathrm{4}!\mathrm{2}!}×\mathrm{3}!=\mathrm{83}\:\mathrm{160} \\ $$$$\mathrm{6}+\mathrm{3}+\mathrm{3}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{6}!\left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{3}!=\mathrm{55}\:\mathrm{440} \\ $$$$\mathrm{5}+\mathrm{5}+\mathrm{2}\:\Rightarrow\frac{\mathrm{12}!}{\left(\mathrm{5}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{3}!=\mathrm{49896} \\ $$$$\mathrm{5}+\mathrm{4}+\mathrm{3}\:\Rightarrow\frac{\mathrm{12}!}{\mathrm{5}!\mathrm{4}!\mathrm{3}!}×\mathrm{3}!=\mathrm{166}\:\mathrm{320} \\ $$$${totally}:\:\:\mathrm{496}\:\mathrm{791} \\ $$ | ||
Commented by SLVR last updated on 02/Nov/22 | ||
$${Now}...{i}\:{got}\:{the}\:{clarity}..{so} \\ $$$${kind}\:{of}\:{sir}..{thanks}\:{a}\:{lot} \\ $$ | ||
Commented by mr W last updated on 03/Nov/22 | ||
$${generating}\:{functions}\:{are}\:{powerful} \\ $$$${tools}\:{for}\:{solving}\:{such}\:{questions}. \\ $$$${we}\:{don}'{t}\:{need}\:{to}\:{enumerate}\:{the} \\ $$$${concrete}\:{combinations}\:{as}\:{i}\:{did}\:{above}. \\ $$$${examples}: \\ $$$$\left.\mathrm{1}\right) \\ $$$${coef}.\:{of}\:{term}\:{x}^{\mathrm{12}} \:{in} \\ $$$$\mathrm{12}!×\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}×\left({e}^{{x}} \right)^{\mathrm{2}} \\ $$$${which}\:{is}\:\mathrm{101}\:\mathrm{376}. \\ $$ | ||
Commented by mr W last updated on 03/Nov/22 | ||
Commented by mr W last updated on 03/Nov/22 | ||
$$\left.\mathrm{2}{a}\right) \\ $$$${coef}.\:{of}\:{term}\:{x}^{\mathrm{12}} \:{in} \\ $$$$\mathrm{3}×\mathrm{12}!×\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}×\left({e}^{{x}} −\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}\right)^{\mathrm{2}} \\ $$$${which}\:{is}\:\mathrm{204}\:\mathrm{336}. \\ $$ | ||
Commented by mr W last updated on 03/Nov/22 | ||
Commented by mr W last updated on 03/Nov/22 | ||
$$\left.\mathrm{2}{b}\right) \\ $$$${without}\:{restriction}:\:\mathrm{3}^{\mathrm{12}} =\mathrm{531}\:\mathrm{441}\:{ways} \\ $$$${all}\:{boxes}\:{have}\:{less}\:{than}\:\mathrm{5}\:{balls},\:{i}.{e}. \\ $$$${each}\:{has}\:\mathrm{4}\:{balls}:\:\frac{\mathrm{12}!}{\left(\mathrm{4}!\right)^{\mathrm{3}} }=\mathrm{34650}\:{ways} \\ $$$${one}\:{box}\:{has}\:{at}\:{least}\:\mathrm{5}\:{balls}: \\ $$$$\Rightarrow\mathrm{531}\:\mathrm{441}−\mathrm{34}\:\mathrm{650}=\mathrm{496}\:\mathrm{791} \\ $$$$ \\ $$$${or}\:{using}\:{generating}\:{functions}: \\ $$ | ||
Commented by mr W last updated on 03/Nov/22 | ||
Commented by mr W last updated on 03/Nov/22 | ||