Question Number 137943 by john_santu last updated on 08/Apr/21 $${Determine}\:{the}\:{term}\:{independent} \\ $$$${of}\:{x}\:{in}\:{the}\:{expansion}\: \\ $$$$\:\:\:\:\left(\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{1}}\:−\frac{{x}−\mathrm{1}}{{x}−{x}^{\mathrm{1}/\mathrm{2}} }\:\right)^{\mathrm{10}} \:. \\ $$ Answered by EDWIN88 last updated…
Question Number 137745 by bemath last updated on 06/Apr/21 $$ \\ $$Each of the digits 2, 4, 6, and 8 can be used once and…
Question Number 6090 by sanusihammed last updated on 13/Jun/16 Answered by Rasheed Soomro last updated on 13/Jun/16 $$\:^{{x}+\frac{{y}}{\mathrm{2}}} \mathrm{C}_{\mathrm{2}} =\mathrm{21}\:\:\wedge\:\:^{\frac{{x}+{y}}{\mathrm{2}}} \mathrm{P}_{\mathrm{2}} =\mathrm{30} \\ $$$${Let}\:\:{x}+\frac{{y}}{\mathrm{2}}={n},\:\:^{{n}} \mathrm{C}_{\mathrm{2}}…
Question Number 5989 by sanusihammed last updated on 08/Jun/16 $${Evaluate}\:\:\:\mathrm{10}\:×\mathrm{12}\:×\:\mathrm{14}\:×\:\mathrm{16}\:×\:\mathrm{18}\:×\:\mathrm{20}\:\:{into}\:{factorial}\:{form} \\ $$ Answered by prakash jain last updated on 08/Jun/16 $$\mathrm{10}×\mathrm{12}×\mathrm{14}×\mathrm{16}×\mathrm{18}×\mathrm{20} \\ $$$$=\mathrm{2}\left(\mathrm{5}×\mathrm{6}×\mathrm{7}×\mathrm{8}×\mathrm{9}×\mathrm{10}\right)=\frac{\mathrm{2}!\mathrm{10}!}{\mathrm{4}!} \\ $$…
Question Number 137035 by mr W last updated on 29/Mar/21 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:{X}\:=\:\mathrm{1345789026} \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{using}\:\mathrm{every}\:\mathrm{digit}\:\mathrm{from}\:{X},\:\mathrm{with}\:\mathrm{condition}: \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{number}\:{n}\:\:\mathrm{is}\:\mathrm{located}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position}\:\mathrm{of}\:{X},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{created}\:\mathrm{number}\:\mathrm{must}\:\mathrm{not}\:\mathrm{contain} \\ $$$$\mathrm{number}\:{n}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position} \\ $$$$…
Question Number 136872 by I want to learn more last updated on 27/Mar/21 Commented by mr W last updated on 27/Mar/21 $${n}=\mathrm{5} \\ $$ Commented…
Question Number 5444 by 3 last updated on 15/May/16 $$\mathrm{8} \\ $$ Answered by FilupSmith last updated on 15/May/16 $$=\mathrm{7}+\mathrm{1} \\ $$ Answered by Rasheed…
Question Number 136448 by adhigenz last updated on 22/Mar/21 $$\mathrm{Mr}.\mathrm{A}\:\mathrm{wants}\:\mathrm{to}\:\mathrm{deliver}\:\mathrm{7}\:\mathrm{letters}\:\mathrm{to}\:\mathrm{his}\:\mathrm{7}\:\mathrm{friends}\:\mathrm{so}\:\mathrm{that}\:\mathrm{each}\:\mathrm{gets}\:\mathrm{1}\:\mathrm{letter}. \\ $$$$\mathrm{All}\:\mathrm{of}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{written}\:\mathrm{of}\:\mathrm{the}\:\mathrm{addresses}\:\mathrm{of}\:\mathrm{his}\:\mathrm{7}\:\mathrm{friends}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{probbility}\:\mathrm{that}, \\ $$$$\mathrm{3}\:\mathrm{of}\:\mathrm{his}\:\mathrm{friends}\:\mathrm{receive}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{letters}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{4}\:\mathrm{receive}\:\mathrm{the}\:\mathrm{wrong}\:\mathrm{ones}. \\ $$ Answered by mr W last updated on 22/Mar/21 $${p}=\frac{{P}_{\mathrm{3}}…
Question Number 136417 by I want to learn more last updated on 21/Mar/21 Answered by EDWIN88 last updated on 22/Mar/21 $$\mathrm{even}\:\mathrm{number}\:=\:\left\{\mathrm{2},\mathrm{4}\right\}\:,\:\mathrm{odd}\:\mathrm{number}=\left\{\mathrm{1},\mathrm{3},\mathrm{5}\right\} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number}\:\mathrm{on}\:\mathrm{two}\:\mathrm{balls} \\ $$$$\mathrm{have}\:\mathrm{probabability}\:=\:\frac{\mathrm{1}+\mathrm{3}}{\mathrm{C}_{\mathrm{2}}…
Question Number 5283 by sanusihammed last updated on 04/May/16 $${If}\:\:\:\:\:\:\:\:\:\mathrm{8}{C}\mathrm{4}\:=\:\mathrm{8}{Cn}\:\:.\:\:{find}\:{the}\:{value}\:{of}\:{n} \\ $$$$ \\ $$ Commented by Rasheed Soomro last updated on 05/May/16 $${n}=\mathrm{4} \\ $$…