Question Number 121851 by bemath last updated on 12/Nov/20 $${How}\:{many}\:{ways}\:{to}\:{arrange}\:{letters}\: \\ $$$${using}\:{letters}\:{in}\:{the}\:{word}\:{OKINKO} \\ $$$$\left({a}\right)\:\mathrm{1260}\:\:\:\:\:\left({b}\right)\:\mathrm{1160}\:\:\:\:\left({c}\right)\:\mathrm{980} \\ $$$$\left({d}\right)\:\mathrm{880}\:\:\:\:\:\:\:\:\left({e}\right)\:\mathrm{720} \\ $$ Commented by liberty last updated on 12/Nov/20…
Question Number 121829 by bemath last updated on 12/Nov/20 Answered by mr W last updated on 12/Nov/20 $${no}\:{empty}:\:\mathrm{3}!=\mathrm{6} \\ $$$${one}\:{empty}:\:\mathrm{3}×\mathrm{2}×\mathrm{3}=\mathrm{18} \\ $$$$\Rightarrow{totally}\:\mathrm{24} \\ $$ Commented…
Question Number 56212 by pieroo last updated on 12/Mar/19 $$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{4}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{3}\:\mathrm{girls} \\ $$$$\mathrm{stand}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{a}.\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{restrictions} \\ $$$$\mathrm{b}.\:\mathrm{if}\:\mathrm{the}\:\mathrm{boys}\:\mathrm{stand}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 12/Mar/19…
Question Number 56025 by pieroo last updated on 08/Mar/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{a}\:\mathrm{student} \\ $$$$\mathrm{arranging}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word} \\ $$$$\mathrm{MATHEMATICS}\:\mathrm{will}\:\mathrm{make}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{vowels}\:\mathrm{be}\:\mathrm{together}\:\mathrm{in}\:\mathrm{any}\:\mathrm{arrangement} \\ $$$$\mathrm{he}\:\mathrm{or}\:\mathrm{she}\:\mathrm{does}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 55502 by mr W last updated on 25/Feb/19 Commented by mr W last updated on 25/Feb/19 $${as}\:{Q}\mathrm{55468},\:{but}\:{passwords}\:{with}\:\mathrm{6} \\ $$$${characters}. \\ $$ Answered by…
Question Number 55468 by peter frank last updated on 24/Feb/19 Answered by mr W last updated on 25/Feb/19 $${to}\:{choose}\:\mathrm{4}\:{characters}\:{there}\:{are}\:{three} \\ $$$${cases}: \\ $$$${case}\:\mathrm{1}:\:{two}\:{letters},\:{one}\:{digit},\:{one}\:{symbol} \\ $$$${to}\:{choose}\:\mathrm{2}\:{letters}\:{there}\:{are}\:{C}_{\mathrm{2}}…
Question Number 120954 by cantor last updated on 04/Nov/20 $$\boldsymbol{{shown}}\:\boldsymbol{{that}} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{n}}!=\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\left(−\mathrm{1}\right)^{\boldsymbol{{k}}} \left(_{\:\boldsymbol{{k}}} ^{\:\boldsymbol{{n}}} \right)\left(\boldsymbol{{n}}−\boldsymbol{{k}}\right)^{\boldsymbol{{n}}} \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{help}}\:\boldsymbol{{me}} \\ $$ Terms of…
Question Number 55192 by Learner last updated on 19/Feb/19 $${what}\:{will}\:{be}\:{the}\:{numbers}\:{of}\:{zeroes}\:{in}\:{the}\:{expension}\:{of} \\ $$$$\left.{a}\right)\:\mathrm{100}!×\mathrm{25}! \\ $$$$\left.{b}\right)\:\mathrm{100}!+\mathrm{25}! \\ $$$$ \\ $$$${please}\:{help} \\ $$ Commented by MJS last updated…
Question Number 55119 by pieroo last updated on 17/Feb/19 $$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers},\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5},\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{made}\:\mathrm{with}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{2},\mathrm{3},\mathrm{4}\:\mathrm{and}\:\mathrm{5}\:\mathrm{where} \\ $$$$\mathrm{no}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{being}\:\mathrm{used}\:\mathrm{more}\:\mathrm{than}\:\mathrm{once} \\ $$$$\mathrm{in}\:\mathrm{each}\:\mathrm{number}? \\ $$ Commented by Tawa1 last updated on 17/Feb/19…
Question Number 55115 by aseerimad last updated on 17/Feb/19 Commented by aseerimad last updated on 17/Feb/19 please help.... Commented by mr W last updated on 17/Feb/19…