Question Number 124806 by bramlexs22 last updated on 06/Dec/20 Answered by mr W last updated on 06/Dec/20 $$\left({i}\right) \\ $$$$\mathrm{6}=\mathrm{1}+\mathrm{5}=\mathrm{2}+\mathrm{4}=\mathrm{3}+\mathrm{3} \\ $$$$\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{6}} ×\mathrm{4}!+{C}_{\mathrm{2}} ^{\mathrm{6}}…
Question Number 190260 by alcohol last updated on 30/Mar/23 $${f}\::\:\left[\mathrm{1},\:\mathrm{3}\right]\:\rightarrow\mathbb{R}\:,\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}} \\ $$$${A}\left(\mathrm{1},\:\mathrm{1}\right) \\ $$$${B}\left(\mathrm{1},\:\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${B}'\left({b},\:\frac{\mathrm{1}}{{b}}\right)\:,\:{b}\:\geqslant\:\mathrm{1} \\ $$$${Find} \\ $$$${i}.\:{equation}\:{of}\:{line}\:{AB}' \\ $$$${ii}.\:{equation}\:{of}\:{tangent}\:{T}\:'\:{to}\:{C}_{{f}} \:{at}\:{point} \\ $$$${with}\:{x}\:=\:\frac{\mathrm{1}\:+\:{b}}{\mathrm{2}}…
Question Number 124712 by benjo_mathlover last updated on 05/Dec/20 $${In}\:{how}\:{many}\:{ways}\:\:{can}\:\mathrm{5}\:{boys}\:{and}\:\mathrm{3}\:{girls} \\ $$$${be}\:{seated}\:{around}\:{a}\:{table}\:{if}\: \\ $$$$\left({i}\right)\:{boy}\:{B}_{\mathrm{3}} \:{and}\:{G}_{\mathrm{2}} \:{are}\:{not}\:{adjacent} \\ $$$$\left({ii}\right)\:{no}\:{girls}\:{are}\:{adjacent}\: \\ $$ Answered by liberty last updated…
Question Number 124709 by benjo_mathlover last updated on 05/Dec/20 $${Between}\:\mathrm{20000}\:{and}\:\mathrm{70000}\: \\ $$$${find}\:{the}\:{number}\:{of}\:{even}\:{integers} \\ $$$${in}\:{which}\:{no}\:{digits}\:{is}\:{repeated} \\ $$ Answered by liberty last updated on 05/Dec/20 $${Let}\:{abcde}\:{be}\:{a}\:{required}\:{even}\:{integer}.\: \\…
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Question Number 59021 by mr W last updated on 03/May/19 $$\left[{Q}\mathrm{58885}\:{reposted}\right] \\ $$$${How}\:{many}\:\mathrm{4}−{digit}\:{numbers}\:{abcd}\:{exist} \\ $$$${which}\:{are}\:{divisible}\:{by}\:\mathrm{3}\:{and}\:{satisfy} \\ $$$${a}\leqslant{b}\leqslant{c}\leqslant{d}? \\ $$ Commented by tanmay last updated on…
Question Number 124400 by mr W last updated on 03/Dec/20 $${How}\:{many}\:{six}-{digit}\:{numbers}\:{contain} \\ $$$${exactly}\:{three}\:{different}\:{digits}? \\ $$ Answered by benjo_mathlover last updated on 03/Dec/20 $${XXXWYZ}\:=\:\mathrm{9}×{C}_{\:\mathrm{3}} ^{\:\mathrm{9}} \:×\:\frac{\mathrm{6}!}{\mathrm{3}!}…
Question Number 124372 by mr W last updated on 02/Dec/20 $${How}\:{many}\:{six}-{digit}\:{numbers}\:{contain} \\ $$$${exactly}\:{six}\:{different}\:{digits}? \\ $$ Answered by bemath last updated on 02/Dec/20 $$=\:\mathrm{9}×{P}_{\mathrm{5}} ^{\:\mathrm{9}} \:=\:\mathrm{9}×\frac{\mathrm{9}!}{\mathrm{4}!}…
Question Number 124237 by mr W last updated on 01/Dec/20 $${How}\:{many}\:\mathrm{6}\:{digit}\:{odd}\:{numbers}\:{have} \\ $$$${different}\:{digits}? \\ $$ Commented by mr W last updated on 02/Dec/20 $$\mathrm{5}×\left({P}_{\mathrm{5}} ^{\mathrm{9}}…
Question Number 124173 by bramlexs22 last updated on 01/Dec/20 Answered by bobhans last updated on 01/Dec/20 $$\left(\mathrm{1}+{kx}\right)\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{5}} \:=\:\left(\mathrm{1}+{kx}\right)\left[\underset{{n}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}{C}_{{n}} ^{\:\mathrm{5}} \left(−\mathrm{2}{x}\right)^{\mathrm{5}−{n}} \:\right] \\ $$$$=\left(\mathrm{1}+{kx}\right)\left({C}_{\mathrm{3}}…