Question Number 160980 by cortano last updated on 10/Dec/21 $$\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{50}\:\mathrm{people} \\ $$$$\:\mathrm{be}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{3}\:\mathrm{groups},\:\mathrm{so} \\ $$$$\:\mathrm{that}\:\mathrm{each}\:\mathrm{group}\:\mathrm{contains}\:\mathrm{members} \\ $$$$\:\mathrm{equal}\:\mathrm{to}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}? \\ $$ Answered by Rasheed.Sindhi last updated on 10/Dec/21…
Question Number 95436 by 675480065 last updated on 25/May/20 $$\int\frac{\mathrm{x}^{\mathrm{4}} \mathrm{dx}}{\mathrm{x}^{\mathrm{8}} +\mathrm{x}^{\mathrm{4}} +\mathrm{1}} \\ $$ Answered by Mr.D.N. last updated on 25/May/20 $$\:\:\:\:\mathrm{I}=\:\int\:\frac{\mathrm{x}^{\mathrm{4}} \mathrm{dx}}{\mathrm{x}^{\mathrm{8}} +\mathrm{x}^{\mathrm{4}}…
Question Number 160829 by cortano last updated on 07/Dec/21 Commented by blackmamba last updated on 07/Dec/21 $$\left(\mathrm{6}−\mathrm{1}\right)!×\mathrm{6}×\mathrm{5}×\mathrm{4}\:=\:\mathrm{120}×\mathrm{120}=\mathrm{14400} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 160665 by mr W last updated on 04/Dec/21 $${In}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{divide}\:\mathrm{20} \\ $$$${students}\:{into}\:\mathrm{5}\:{groups}\:{with}\:\mathrm{4}\:{students} \\ $$$${in}\:{each}\:{group}\:{such}\:{that}\:{any}\:{two} \\ $$$${students}\:{don}'{t}\:{meet}\:{each}\:{other}\:{in} \\ $$$${more}\:{than}\:{one}\:{group}. \\ $$ Commented by talminator2856791 last…
Question Number 160389 by mr W last updated on 28/Nov/21 Commented by mr W last updated on 29/Nov/21 $${thanks}\:{for}\:{trying}! \\ $$$${but}\:{it}'{s}\:{wrong}\:{sir}.\:{i}\:{think}\:{you}\: \\ $$$${misunderstood}\:{the}\:{question}.\:{it}\:{is}\:{not} \\ $$$${to}\:{select}\:{one}\:{pair}\:{from}\:\mathrm{20}\:{people},\:{but}…
Question Number 94397 by peter frank last updated on 18/May/20 Answered by mr W last updated on 18/May/20 $$\left(\mathrm{3}{x}+\mathrm{5}\right)^{\mathrm{10}} \\ $$$$=\mathrm{5}^{\mathrm{10}} \left(\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{5}}\right)^{\mathrm{10}} \\ $$$$=\mathrm{5}^{\mathrm{10}} \underset{{k}=\mathrm{0}}…
Question Number 94352 by i jagooll last updated on 18/May/20 Commented by mr W last updated on 18/May/20 $$\frac{\mathrm{2}×{C}_{\mathrm{3}} ^{\mathrm{10}} ×\mathrm{3}!×\mathrm{7}!}{\mathrm{11}!}=\frac{\mathrm{2}}{\mathrm{11}} \\ $$ Commented by…
Question Number 159715 by cortano last updated on 20/Nov/21 Commented by mr W last updated on 20/Nov/21 $${if}\:{one}\:{draws}\:\mathrm{12}\:{balls},\:{they}\:{could}\:{be} \\ $$$$\mathrm{3}\:{red}\:{ones},\:\mathrm{3}\:{green}\:{ones},\:\mathrm{3}\:{blacks}, \\ $$$$\mathrm{3}\:{blue}\:{one}.\:{if}\:{one}\:{draws}\:{one}\:{ball}\:{more}, \\ $$$${one}\:{color}\:{must}\:{have}\:\mathrm{4}\:{balls}.\:{so}\:{the} \\…
Question Number 93764 by DuDono last updated on 14/May/20 $$\mathrm{define} \\ $$$$\underset{{x}=\mathrm{0}} {\overset{{k}} {\sum}}\frac{{n}!}{{x}!\left({n}−{x}\right)!} \\ $$$$\mathrm{without}\:\Sigma\:\mathrm{symbol} \\ $$ Commented by mr W last updated on…
Question Number 159259 by pticantor last updated on 14/Nov/21 $$\boldsymbol{{montre}}\:\boldsymbol{{que}}\: \\ $$$$\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{C}}_{\mathrm{2}\boldsymbol{{n}}} ^{\mathrm{2}\boldsymbol{{k}}} =\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}} \\ $$ Answered by mr W last updated on…