Category: Permutation and Combination

Question Number 87074 by jagoll last updated on 02/Apr/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{t}^{\mathrm{3}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{expanssion}\:\left\{\frac{\mathrm{1}−\mathrm{t}^{\mathrm{6}} }{\mathrm{1}−\mathrm{t}}\right\}^{\mathrm{3}} \: \\ $$ Commented by mr W last updated on 02/Apr/20 $${t}^{\mathrm{2}}…

Question Number 21405 by Tinkutara last updated on 23/Sep/17 $$\mathrm{Four}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{rolled}.\:\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways} \\ $$$$\mathrm{in}\:\mathrm{which}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{die}\:\mathrm{shows}\:\mathrm{3},\:\mathrm{is} \\ $$ Answered by myintkhaing last updated on 23/Sep/17 $$\mathrm{1}−\left(\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{5}}{\mathrm{6}}\right)=\frac{\mathrm{671}}{\mathrm{1296}} \\ $$ Commented…

Question Number 21404 by Tinkutara last updated on 23/Sep/17 $$\mathrm{Prove}\:\mathrm{that}\:\left(\mathrm{6}{n}\right)!\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}^{\mathrm{2}{n}} .\mathrm{3}^{{n}} . \\ $$ Answered by dioph last updated on 23/Sep/17 $$\mathrm{if}\:{n}\:>\:\mathrm{1},\:\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least}\:\mathrm{3}{n}\:\mathrm{even} \\ $$$$\mathrm{numbers}\:\mathrm{and}\:\mathrm{2}{n}\:\mathrm{multiples}\:\mathrm{of}\:\mathrm{3}\:\mathrm{in} \\…

Question Number 86684 by john santu last updated on 30/Mar/20 $$\mathrm{If}\:\left(\mathrm{1}+\mathrm{px}+\mathrm{qx}^{\mathrm{2}} \right)^{\mathrm{8}} \:=\:\mathrm{1}+\mathrm{8x}+\mathrm{52x}^{\mathrm{2}} +\mathrm{kx}^{\mathrm{3}} +… \\ $$$$\mathrm{find}\:\mathrm{p}\:,\:\mathrm{q}\:\mathrm{and}\:\mathrm{k}.\: \\ $$ Answered by jagoll last updated on…

Question Number 151832 by DELETED last updated on 23/Aug/21 Answered by DELETED last updated on 23/Aug/21 $$\left.\mathrm{57}\right).\:\mathrm{R}_{\mathrm{s}} =\mathrm{2R}+\mathrm{2R}=\mathrm{4R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }=\frac{\mathrm{1}}{\mathrm{2R}}+\frac{\mathrm{1}}{\mathrm{4R}}=\frac{\mathrm{3}}{\mathrm{4R}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\mathrm{p}} =\frac{\mathrm{4R}}{\mathrm{3}}\: \\…