Question Number 163443 by henderson last updated on 07/Jan/22 $$\mathrm{p}\:\leqslant\:\mathrm{n}\: \\ $$$$\mathrm{find}\:\:\frac{\mathrm{A}_{\mathrm{n}} ^{\mathrm{p}} }{\mathrm{A}_{\mathrm{n}−\mathrm{1}} ^{\mathrm{p}} }. \\ $$ Commented by mr W last updated on…
Question Number 163313 by SLVR last updated on 06/Jan/22 $${Find}\:{the}\:{non}\:{negative}\:{integer} \\ $$$${solutions}\:{of}\:\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{5}{z}=\mathrm{60} \\ $$ Answered by mr W last updated on 06/Jan/22 $${let}\:\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{5}{u} \\ $$$$\mathrm{5}{u}+\mathrm{5}{z}=\mathrm{60}…
Question Number 163283 by greg_ed last updated on 05/Jan/22 $$\mathrm{hi}\:! \\ $$We store 5 objects in three discernible drawers. Suppose that the different ways of…
Question Number 163263 by Ar Brandon last updated on 05/Jan/22 Answered by qaz last updated on 05/Jan/22 $$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{k}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}=\frac{\partial}{\partial\mathrm{x}}\mid_{\mathrm{x}=\mathrm{1}} \left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}\left(−\mathrm{x}\right)^{\mathrm{k}} −\mathrm{1}\right)=\frac{\partial}{\partial\mathrm{x}}\mid_{\mathrm{x}=\mathrm{1}}…
Question Number 32093 by NECx last updated on 19/Mar/18 $${Find}\:{the}\:{number}\:{of}\:{ways}\:{of} \\ $$$${selecting}\:\mathrm{9}\:{balls}\:{from}\:\mathrm{6}\:{red}\:{balls}, \\ $$$$\mathrm{5}\:{white}\:{balls}\:{and}\:\mathrm{5}\:{blue}\:{balls}\:{if} \\ $$$${each}\:{selection}\:{consists}\:{of}\:\mathrm{3}\:{balls} \\ $$$${of}\:{each}\:{colour}. \\ $$ Answered by Joel578 last updated…
Question Number 97600 by Power last updated on 08/Jun/20 Answered by mahdi last updated on 08/Jun/20 $$\mathrm{19}\:\mathrm{trangles} \\ $$$$\mathrm{3},\mathrm{4},\mathrm{5}\:\&\:\mathrm{3},\mathrm{4},\mathrm{6}\:\&\:\mathrm{4},\mathrm{5},\mathrm{6}\:\&\:\mathrm{3},\mathrm{5},\mathrm{6} \\ $$$$\mathrm{3},\mathrm{3},\mathrm{4}\:\&\:\mathrm{3},\mathrm{3},\mathrm{5}\:\&\:\mathrm{4},\mathrm{4},\mathrm{3}\:\&\:\mathrm{4},\mathrm{4},\mathrm{5}\:\&\:\mathrm{4},\mathrm{4},\mathrm{6} \\ $$$$\mathrm{5},\mathrm{5},\mathrm{3}\:\&\:\mathrm{5},\mathrm{5},\mathrm{4}\:\&\:\mathrm{5},\mathrm{5},\mathrm{6}\:\&\:\mathrm{6},\mathrm{6},\mathrm{3}\:\&\:\mathrm{6},\mathrm{6},\mathrm{4}\:\&\:\mathrm{6},\mathrm{6},\mathrm{5} \\ $$$$\mathrm{3},\mathrm{3},\mathrm{3}\:\&\:\mathrm{4},\mathrm{4},\mathrm{4}\:\&\:\mathrm{5},\mathrm{5},\mathrm{5}\:\&\:\mathrm{6},\mathrm{6},\mathrm{6}…
Question Number 97492 by 675480065 last updated on 08/Jun/20 Answered by smridha last updated on 08/Jun/20 $$\boldsymbol{{let}}\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)=−\boldsymbol{{k}}\:\boldsymbol{{so}}\:\boldsymbol{{we}}\:\boldsymbol{{get}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\mathrm{2021}\boldsymbol{{k}}} \boldsymbol{{k}}^{\mathrm{2020}} \boldsymbol{{dk}} \\ $$$$=\frac{\boldsymbol{\Gamma}\left(\mathrm{2021}\right)}{\left(\mathrm{2021}\right)^{\mathrm{2021}}…
Question Number 31714 by gunawan last updated on 13/Mar/18 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{1}} {\overset{{n}−\mathrm{2}} {\sum}}{k}\begin{pmatrix}{\:{n}−\mathrm{2}}\\{\:\:{k}}\end{pmatrix}\begin{pmatrix}{{n}+\mathrm{2}}\\{{k}+\mathrm{2}}\end{pmatrix}\:=\:\left({n}−\mathrm{2}\right)\begin{pmatrix}{\mathrm{2}{n}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31713 by gunawan last updated on 13/Mar/18 $$\mathrm{Given}\:{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({n}+\mathrm{1}−{k}\right)=\begin{pmatrix}{\:{n}+\mathrm{2}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix} \\ $$ Commented by abdo imad last updated on…
Question Number 97045 by john santu last updated on 06/Jun/20 $$\mathrm{a}\:\mathrm{committee}\:\mathrm{consisting}\:\mathrm{of} \\ $$$$\mathrm{5}\:\mathrm{men}\:\&\:\mathrm{4}\:\mathrm{women}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\: \\ $$$$\mathrm{chosen}\:\mathrm{from}\:\mathrm{8}\:\mathrm{men}\:\&\:\mathrm{4}\:\mathrm{women}. \\ $$$$\mathrm{If}\:\mathrm{one}\:\mathrm{man}\:\&\:\mathrm{one}\:\mathrm{woman}\:\mathrm{are}\:\mathrm{husband}\: \\ $$$$\&\:\mathrm{wife}\:,\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\: \\ $$$$\mathrm{the}\:\mathrm{committee}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{if} \\ $$$$\mathrm{only}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{husband}\:\&\:\mathrm{wife} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{chosen}\:?…