Question Number 124400 by mr W last updated on 03/Dec/20 $${How}\:{many}\:{six}-{digit}\:{numbers}\:{contain} \\ $$$${exactly}\:{three}\:{different}\:{digits}? \\ $$ Answered by benjo_mathlover last updated on 03/Dec/20 $${XXXWYZ}\:=\:\mathrm{9}×{C}_{\:\mathrm{3}} ^{\:\mathrm{9}} \:×\:\frac{\mathrm{6}!}{\mathrm{3}!}…
Question Number 124372 by mr W last updated on 02/Dec/20 $${How}\:{many}\:{six}-{digit}\:{numbers}\:{contain} \\ $$$${exactly}\:{six}\:{different}\:{digits}? \\ $$ Answered by bemath last updated on 02/Dec/20 $$=\:\mathrm{9}×{P}_{\mathrm{5}} ^{\:\mathrm{9}} \:=\:\mathrm{9}×\frac{\mathrm{9}!}{\mathrm{4}!}…
Question Number 124237 by mr W last updated on 01/Dec/20 $${How}\:{many}\:\mathrm{6}\:{digit}\:{odd}\:{numbers}\:{have} \\ $$$${different}\:{digits}? \\ $$ Commented by mr W last updated on 02/Dec/20 $$\mathrm{5}×\left({P}_{\mathrm{5}} ^{\mathrm{9}}…
Question Number 124173 by bramlexs22 last updated on 01/Dec/20 Answered by bobhans last updated on 01/Dec/20 $$\left(\mathrm{1}+{kx}\right)\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{5}} \:=\:\left(\mathrm{1}+{kx}\right)\left[\underset{{n}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{{n}} ^{\:\mathrm{5}} \left(−\mathrm{2}{x}\right)^{\mathrm{5}−{n}} \:\right] \\ $$$$=\left(\mathrm{1}+{kx}\right)\left({C}_{\mathrm{3}}…
Question Number 189614 by Rupesh123 last updated on 19/Mar/23 Answered by Frix last updated on 19/Mar/23 $${a}=\mathrm{2}{m}−\mathrm{1}\wedge\mathrm{1}\leqslant{m}\leqslant\mathrm{14}\wedge{b}=\mathrm{2}{n}−\mathrm{1}\wedge{m}+\mathrm{1}\leqslant{n}\leqslant\mathrm{15} \\ $$$${a}=\mathrm{2}{m}\wedge\mathrm{1}\leqslant{m}\leqslant\mathrm{14}\wedge{b}=\mathrm{2}{n}\wedge{m}+\mathrm{1}\leqslant{n}\leqslant\mathrm{15} \\ $$$$\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{14}} {\sum}}{k}=\mathrm{210} \\ $$…
Question Number 58373 by Tawa1 last updated on 22/Apr/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\:\mathrm{x}^{\mathrm{6}} \:\:\mathrm{in}\:\:\:\left(\mathrm{2x}\:+\:\mathrm{1}\right)^{\mathrm{6}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$ Commented by maxmathsup by imad last updated on 24/Apr/19 $${we}\:{have}\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{6}}…
Question Number 123878 by liberty last updated on 29/Nov/20 $${Nine}\:{chairs}\:{in}\:{a}\:{row}\:{are}\:{to}\:{be}\:{occupied}\:{by} \\ $$$${six}\:{students}\:{and}\:{Prof}\:{George}\:,\:{Prof}\:{Pieter}, \\ $$$${and}\:{Prof}\:{John}.\:{These}\:{three}\:{professors}\:{arrive} \\ $$$${before}\:{the}\:{six}\:{students}\:{and}\:{decide}\:{to}\:{choose} \\ $$$${their}\:{chairs}\:{so}\:{that}\:{each}\:{professor}\:{will}\:{be}\:{between} \\ $$$${two}\:{students}.\:{In}\:{how}\:{many}\:{ways}\:{can}\:{Professor}\:{George} \\ $$$$,\:{Pieter}\:{and}\:{John}\:{choose}\:{their}\:{chairs}\:?\: \\ $$ Answered…
Question Number 123695 by pipin last updated on 27/Nov/20 $$\boldsymbol{\mathrm{There}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{many}}\:\boldsymbol{\mathrm{ways}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{answer}} \\ $$$$\mathrm{5}\:\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{of}}\:\mathrm{7}\:\boldsymbol{\mathrm{multiple}}\:\boldsymbol{\mathrm{choice}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{with}} \\ $$$$\mathrm{5}\:\boldsymbol{\mathrm{choices}}? \\ $$ Answered by mr W last updated on 27/Nov/20 $${C}_{\mathrm{5}}…
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Question Number 123577 by bramlexs22 last updated on 26/Nov/20 Answered by Olaf last updated on 26/Nov/20 $$\begin{cases}{\frac{{g}}{{r}+{g}}\:=\:\frac{\mathrm{4}}{\mathrm{9}}\:\left(\mathrm{1}\right)}\\{\frac{{g}+\mathrm{2}}{\left({r}+\mathrm{4}\right)+\left({g}+\mathrm{2}\right)}\:=\:\frac{\mathrm{10}}{\mathrm{23}}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)\::\:{r}+{g}\:=\:\frac{\mathrm{9}}{\mathrm{4}}{g}\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\::\:\frac{{g}+\mathrm{2}}{\frac{\mathrm{9}}{\mathrm{4}}{g}+\mathrm{6}}\:\:=\:\frac{\mathrm{10}}{\mathrm{23}} \\ $$$${g}+\mathrm{2}\:=\:\frac{\mathrm{10}}{\mathrm{23}}\left(\frac{\mathrm{9}}{\mathrm{4}}{g}+\mathrm{6}\right) \\ $$$${g}\left(\mathrm{1}−\frac{\mathrm{45}}{\mathrm{46}}\right)\:=\:\frac{\mathrm{60}}{\mathrm{23}}−\mathrm{2}\:=\:\frac{\mathrm{14}}{\mathrm{23}}…