Question Number 105265 by bemath last updated on 27/Jul/20 Answered by Cinezoidy1 last updated on 27/Jul/20 $${P}\left({F}_{\mathrm{5}} \cup{V}\right)={P}\left({F}_{\mathrm{5}} \right)+{P}\left({V}\right)−{P}\left({F}_{\mathrm{5}} \cap{V}\right) \\ $$$${P}\left({F}_{\mathrm{5}} \cap{V}\right)=\frac{\mathrm{5}{letters}}{\mathrm{26}}+\frac{\mathrm{5}{vowels}}{\mathrm{26}}−\frac{\mathrm{2}\left({a}\:{and}\:{e}\right)}{\mathrm{26}} \\ $$$${P}\left({F}_{\mathrm{5}}…
Question Number 170794 by mr W last updated on 30/May/22 $$\mathrm{2}{n}\:\mathrm{objects}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{three}\:\mathrm{kinds}\:\mathrm{are} \\ $$$$\mathrm{given}\:\mathrm{to}\:\mathrm{two}\:\mathrm{persons},\:\mathrm{so}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{person}\:\mathrm{gets}\:\mathrm{3}{n}\:\mathrm{objects}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{1}\:\mathrm{ways}. \\ $$ Answered by aleks041103 last updated…
Question Number 104370 by bemath last updated on 21/Jul/20 Answered by john santu last updated on 21/Jul/20 $$\Box\bigtriangleup\Box\bigtriangleup \\ $$$$\Box={letter} \\ $$$$\bigtriangleup={number} \\ $$$${password}\:=\:\mathrm{4}×\mathrm{4}×\mathrm{1}×\mathrm{6}\:=\:\mathrm{96}\:. \\…
Question Number 104155 by mr W last updated on 19/Jul/20 Commented by mr W last updated on 19/Jul/20 $${an}\:{unsolved}\:{old}\:{question} \\ $$ Commented by mr W…
Question Number 104134 by bemath last updated on 19/Jul/20 $${what}\:{is}\:{the}\:{coefficient}\:{x}^{\mathrm{15}} \\ $$$${in}\:{the}\:{expansion}\:{of}\:{x}^{\mathrm{6}} \left(\mathrm{1}−{x}\right)^{\mathrm{11}} \\ $$ Answered by bobhans last updated on 19/Jul/20 $$''\boldsymbol{{x}}^{\mathrm{6}} \left(\mathrm{1}−\boldsymbol{{x}}\right)^{\mathrm{11}} \:=\:\boldsymbol{{x}}^{\mathrm{6}}…
Question Number 103903 by mr W last updated on 18/Jul/20 $${a}\:\mathrm{100}\:{cm}\:{long}\:{rod}\:{should}\:{be}\:{divided} \\ $$$${into}\:\mathrm{3}\:{parts}.\:{the}\:{length}\:{of}\:{each}\:{part} \\ $$$${in}\:{cm}\:{should}\:{be}\:{integer}.\:{in}\:{how}\: \\ $$$${many}\:{different}\:{ways}\:{can}\:{this}\:{be} \\ $$$${done}? \\ $$ Commented by bobhans last…
Question Number 103826 by bobhans last updated on 17/Jul/20 $${In}\:{the}\:{expansion}\:{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{20}} \:{if}\:{the} \\ $$$${coefficient}\:{of}\:{x}^{{r}} \:{is}\:{twice}\:{the}\:{coefficient} \\ $$$${of}\:{x}^{{r}−\mathrm{1}} ,\:{what}\:{the}\:{value}\:{of}\:{the} \\ $$$${coefficient}?\: \\ $$ Answered by bramlex last…
Question Number 103716 by mr W last updated on 16/Jul/20 Commented by mr W last updated on 16/Jul/20 $${after}\:{thinking}\:{about}\:{it}\:{for}\:{two}\:{days}, \\ $$$${i}\:{think}\:{i}\:{have}\:{found}\:{the}\:{general} \\ $$$${solution}\:{which}\:{is}\:{simply} \\ $$$$\boldsymbol{{C}}_{\boldsymbol{{m}}}…
Question Number 169159 by SLVR last updated on 25/Apr/22 $${The}\:{remainder}\:{of} \\ $$$$\mathrm{13}^{\mathrm{13}} {when}\: \\ $$$${divided}\:{by}\:\mathrm{99}\:{is} \\ $$ Commented by SLVR last updated on 25/Apr/22 $${kindly}\:{solve}..…
Question Number 103606 by bemath last updated on 16/Jul/20 $${an}\:{integer}\:{n}\:{between}\:\mathrm{1}\:{and}\:\mathrm{98}\:, \\ $$$${inclusive}\:{is}\:{to}\:{be}\:{chosen}\:{at} \\ $$$${random}.\:{what}\:{is}\:{the}\:{probability} \\ $$$${that}\:{n}\left({n}+\mathrm{1}\right)\:{will}\:{be}\:{divisible}\:{by}\:\mathrm{3} \\ $$ Commented by bemath last updated on 16/Jul/20…