Question Number 112531 by Aina Samuel Temidayo last updated on 08/Sep/20 $$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{cardboard}\:\mathrm{is}\:\mathrm{8cm}\:\mathrm{long} \\ $$$$\mathrm{and}\:\mathrm{6cm}\:\mathrm{wide}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{least} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{beads}\:\mathrm{you}\:\mathrm{can}\:\mathrm{arrange}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{board}\:\mathrm{such}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{beads}\:\mathrm{that}\:\mathrm{are}\:\mathrm{less}\:\mathrm{than} \\ $$$$\sqrt{\mathrm{10}}\mathrm{cm}\:\mathrm{apart}. \\ $$ Commented…
Question Number 112195 by 675480065 last updated on 06/Sep/20 $$\underset{\mathrm{n}=\mathrm{1}\:} {\overset{\mathrm{11}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \left(\mathrm{4n}+\mathrm{2}\right)}{\mathrm{4n}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\mathrm{please}\:\mathrm{help} \\ $$ Answered by MJS_new last updated on 06/Sep/20 $$\mathrm{simply}\:\mathrm{calculate}\:\mathrm{it}!\:\mathrm{it}'\mathrm{s}\:\mathrm{only}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{11}…
Question Number 177597 by cortano1 last updated on 07/Oct/22 Answered by mr W last updated on 07/Oct/22 $$\boldsymbol{{method}}\:\mathrm{1}:\:\boldsymbol{{without}}\:\boldsymbol{{using}}\:\boldsymbol{{formula}} \\ $$$$\mathrm{1}×\mathrm{1}×{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{2}^{\mathrm{3}} \\ $$$$+\left[\mathrm{1}×\left(−\mathrm{1}\right)+\mathrm{3}×\mathrm{1}\right]×{C}_{\mathrm{2}} ^{\mathrm{6}}…
Question Number 111937 by bemath last updated on 05/Sep/20 Answered by bobhans last updated on 05/Sep/20 $$\:\mathrm{let}\:\mathrm{the}\:\mathrm{first}\:\mathrm{five}\:\mathrm{question}\:\mathrm{is}\:\mathrm{number}\::\: \\ $$$$\mathrm{1};\:\mathrm{2};\:\mathrm{3};\:\mathrm{4};\:\mathrm{5}\: \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\Rightarrow\:\mathrm{C}\:_{\mathrm{4}} ^{\mathrm{5}} \:×\:\mathrm{C}\:_{\mathrm{4}} ^{\mathrm{5}} \:=\:\mathrm{25}\:…
Question Number 111906 by mr W last updated on 05/Sep/20 Answered by PRITHWISH SEN 2 last updated on 22/Sep/20 $$\mathrm{Disarrangement}\:\mathrm{of}\:\mathrm{6}\:\mathrm{cards}\:!\mathrm{6}=\mathrm{265} \\ $$$$\mathrm{now}\:\mathrm{in}\:\mathrm{this}\:\mathrm{265}\:\mathrm{ways}\:\mathrm{card}\:\mathrm{no}.\:\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{in}\: \\ $$$$\mathrm{envelope}\:\mathrm{no}.\:\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\:\mathrm{or}\:\mathrm{6}\:\mathrm{and}\:\mathrm{in}\:\mathrm{each}\:\mathrm{case}\:\mathrm{the} \\…
Question Number 177309 by SLVR last updated on 03/Oct/22 $${How}\:{many}\:\mathrm{3}\:{digited}\:{numbers}\: \\ $$$$\left.{which}\:{are}\:{divisible}\:{by}\:\mathrm{1}\right)\mathrm{3} \\ $$$$\left.\mathrm{2}\left.\right)\left.\mathrm{4}\left.\:\left.\:\left.\:\:\mathrm{3}\right)\mathrm{5}\:\:\:\:\mathrm{4}\right)\mathrm{6}\:\:\:\mathrm{5}\right)\mathrm{7}\:\:\:\:\mathrm{6}\right)\mathrm{8}\:\:\mathrm{7}\right)\mathrm{9} \\ $$$${with}\:{repetetion}\:{of}\:{digits}\:{is} \\ $$$${NOT}\:{allowed}…{one}\:{problem}\:{process} \\ $$ Commented by SLVR last updated…
Question Number 111732 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\mathrm{A}\:\mathrm{blind}\:\mathrm{man}\:\mathrm{is}\:\mathrm{to}\:\mathrm{place}\:\mathrm{5}\:\mathrm{letters}\:\mathrm{into}\:\mathrm{5} \\ $$$$\mathrm{pigeon}\:\mathrm{holes},\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{4}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{letters}\:\mathrm{be}\:\mathrm{wrongly}\:\mathrm{placed}? \\ $$$$\left(\mathrm{note}\:\mathrm{that}\:\mathrm{only}\:\mathrm{one}\:\mathrm{letter}\:\mathrm{must}\:\mathrm{be}\:\mathrm{in}\:\mathrm{a}\right. \\ $$$$\left.\mathrm{pigeon}\:\mathrm{hole}\right) \\ $$ Answered by mr…
Question Number 46143 by aseerimad last updated on 21/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18 $$\mathrm{52}{C}_{\mathrm{26}} ×\mathrm{26}{C}_{\mathrm{26}} \\ $$$$=\frac{\mathrm{52}!}{\mathrm{26}!\mathrm{26}!} \\ $$ Commented by aseerimad…
Question Number 177040 by cortano1 last updated on 30/Sep/22 Answered by mr W last updated on 30/Sep/22 $${total}\:{number}\:{of}\:\mathrm{9}−{digit}\:{numbers}\:{is}: \\ $$$$\mathrm{9}!=\mathrm{362880} \\ $$$$ \\ $$$${number}\:{of}\:{those}\:{which}\:{are}\:{divisible} \\…
Question Number 177008 by Ar Brandon last updated on 29/Sep/22 Answered by Rasheed.Sindhi last updated on 29/Sep/22 $${Let}\:{t}\:{is}\:{total}\:{of}\:{card}\:{numbers}\:{chose}\:{by}\:{players}. \\ $$$${Efe}'{s}\:{card}\:{number}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\mathrm{6}\leqslant{t}\leqslant\mathrm{20} \\ $$$${If}\:{t}\in\mathbb{P}\:{then}\:{possible}\:{values}\:{for}\:{t}: \\…