Question Number 218299 by mr W last updated on 05/Apr/25 $$\mathrm{10}\:{couples}\:{are}\:{invited}\:{to}\:{a}\:{dinner}\:{and} \\ $$$${should}\:{be}\:{seated}\:{at}\:{a}\:{round}\:{table}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{the}\:{host}\:{do} \\ $$$${this}, \\ $$$$\left.\mathrm{1}\right)\:{generally} \\ $$$$\left.\mathrm{2}\right)\:{if}\:{two}\:{men}\:{should}\:{not}\:{sit}\:{next} \\ $$$$\:\:\:\:\:{to}\:{each}\:{other} \\ $$$$\left.\mathrm{3}\left.\right)\:{as}\:\mathrm{2}\right),\:{but}\:{two}\:{speicial}\:{couples}\:…
Question Number 218169 by vile last updated on 31/Mar/25 $${P}\left(\mathrm{5},\mathrm{6}\right)=\frac{\mathrm{15}!}{\left(\mathrm{15}−\mathrm{6}\right)}\:=\:\frac{\mathrm{15}!}{\mathrm{9}!}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{15}×\mathrm{14}×\mathrm{131}×\mathrm{2}×\mathrm{11}×\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{15}×\mathrm{14}×\mathrm{13}×\mathrm{12}×\mathrm{11}×\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3},\mathrm{603},\mathrm{600} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Terms of Service Privacy Policy…
Question Number 218129 by mr W last updated on 30/Mar/25 $${how}\:{many}\:{different}\:{words}\:{can}\:{be} \\ $$$${formed}\:{from}\:{the}\:{word}\: \\ $$$$\boldsymbol{\mathrm{MATHEMATICS}}? \\ $$$${note}:\:\:{here}\:{a}\:{word}\:{should}\:{have}\:{at}\: \\ $$$${least}\:{two}\:{letters},\:{but}\:{mustn}'{t}\:{have}\:{a} \\ $$$${meaning}. \\ $$ Answered by…
Question Number 217402 by mr W last updated on 12/Mar/25 Answered by vnm last updated on 13/Mar/25 $${C}_{\mathrm{5}} ^{\mathrm{2}} \centerdot\mathrm{4}!\centerdot\mathrm{4}=\mathrm{10}\centerdot\mathrm{24}\centerdot\mathrm{4}=\mathrm{960} \\ $$ Commented by mr…
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Question Number 213796 by efronzo1 last updated on 17/Nov/24 Answered by A5T last updated on 17/Nov/24 $${x}_{\mathrm{0}} ={k}\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\Rightarrow{x}_{\mathrm{2}} =\frac{−\mathrm{1}}{{k}}\Rightarrow{x}_{\mathrm{3}} =\frac{{k}−\mathrm{1}}{\mathrm{1}+{k}}\Rightarrow{x}_{\mathrm{4}} =\frac{\mathrm{2}{k}}{\mathrm{2}}={k} \\ $$$$\Rightarrow{x}_{\mathrm{4}{n}} ={k}=\mathrm{2022}…
Question Number 212686 by mr W last updated on 21/Oct/24 $${in}\:{how}\:{many}\:{ways}\:{can}\:{a}\:{teacher} \\ $$$${divide}\:{his}\:\mathrm{10}\:{studens}\:{into}\:\mathrm{4}\:{groups} \\ $$$${such}\:{that}\:{each}\:{group}\:{has}\:{at}\:{least}\:\mathrm{2}\: \\ $$$${students}? \\ $$ Commented by Spillover last updated on…
Question Number 210290 by Adeyemi889 last updated on 05/Aug/24 Answered by mr W last updated on 05/Aug/24 $$\left({i}\right) \\ $$$${C}_{\mathrm{3}} ^{\mathrm{7}} {C}_{\mathrm{3}} ^{\mathrm{6}} =\mathrm{700}\:{ways} \\…
Question Number 209281 by Tawa11 last updated on 06/Jul/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{r},\:\mathrm{if}\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{r}} \:\:=\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{2r}\:\:+\:\:\mathrm{1}} \\ $$ Commented by klipto last updated on 06/Jul/24 $$\:\:\:\:\:\:\:^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{C}}_{\mathrm{r}} =^{\mathrm{n}}…
Question Number 208662 by efronzo1 last updated on 20/Jun/24 $$\:\:\frac{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}\:+\mathrm{3}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{5}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{2}}\end{pmatrix}\:+…+\left(\mathrm{2n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}}{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}\:+\mathrm{2}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{2}}\end{pmatrix}\:+\:\mathrm{3}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{3}}\end{pmatrix}\:+…+\mathrm{n}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}}\:=\frac{\mathrm{23}}{\mathrm{11}} \\ $$$$\:\mathrm{n}=? \\ $$ Answered by Berbere last updated on 20/Jun/24 $${A}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right)\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix};\underset{{k}=\mathrm{0}} {\overset{{n}}…