Question Number 220876 by Spillover last updated on 20/May/25 Answered by Rasheed.Sindhi last updated on 20/May/25 $$\left({a}\right)\:\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n} \\ $$$$\:\:\:\:\:\:=\frac{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}\left({n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:=\frac{\left({n}+\mathrm{2}\right)!}{\left({n}−\mathrm{1}\right)!} \\ $$ Commented by…
Question Number 220877 by Spillover last updated on 20/May/25 Answered by Rasheed.Sindhi last updated on 20/May/25 $$\left({a}\right)\:\left({n}+\mathrm{2}\right)!+\left({n}+\mathrm{1}\right)!+{n}! \\ $$$$=\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}!+\left({n}+\mathrm{1}\right){n}!+{n}! \\ $$$$={n}!\left(\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)+\left({n}+\mathrm{1}\right)+\mathrm{1}\right) \\ $$$$\left.={n}!\left\{\:\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)+\mathrm{1}\right)\right\} \\ $$$$={n}!\left({n}^{\mathrm{2}}…
Question Number 220878 by Spillover last updated on 20/May/25 Answered by Rasheed.Sindhi last updated on 22/May/25 $$\frac{{n}!}{\left({n}−{r}\right)!{r}!}+\frac{\mathrm{2}×{n}!}{\left({n}−{r}+\mathrm{1}\right)!\left({r}−\mathrm{1}\right)!}+\frac{{n}!}{\left({n}−{r}+\mathrm{2}\right)!\left({r}−\mathrm{2}\right)!} \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!{r}\left({r}−\mathrm{1}\right)\left({r}−\mathrm{2}\right)!}+\frac{\mathrm{2}×{n}!}{\left({n}−{r}+\mathrm{1}\right)\left({n}−{r}\right)!\left({r}−\mathrm{1}\right)\left({r}−\mathrm{2}\right)!}+\frac{{n}!}{\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!} \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!}\left(\frac{\mathrm{1}}{{r}\left({r}−\mathrm{1}\right)}+\frac{\mathrm{2}}{\left({n}−{r}+\mathrm{1}\right)\left({r}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)}\right) \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!}\left(\frac{\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)+\mathrm{2}{r}\left({n}−{r}+\mathrm{2}\right)+{r}\left({r}−\mathrm{1}\right)}{{r}\left({r}−\mathrm{1}\right)\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)}\right) \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!}\left(\frac{\left({n}−{r}\right)^{\mathrm{2}}…
Question Number 220872 by Spillover last updated on 20/May/25 Answered by mr W last updated on 21/May/25 $${P}_{{r}} ^{{n}} ={r}!×{C}_{{r}} ^{{n}} =\frac{{n}!}{\left({n}−{r}\right)!} \\ $$ Commented…
Question Number 220873 by Spillover last updated on 20/May/25 Answered by mr W last updated on 20/May/25 $$\mathrm{18}!×\mathrm{19}×\mathrm{18}=\mathrm{18}×\mathrm{19}! \\ $$$${or} \\ $$$$\mathrm{20}!−\mathrm{2}×\mathrm{19}!=\mathrm{18}×\mathrm{19}! \\ $$ Commented…
Question Number 220874 by Spillover last updated on 20/May/25 Commented by Spillover last updated on 20/May/25 Commented by Spillover last updated on 20/May/25 $${its}\:{n}!\:\:{not}\:{n} \\…
Question Number 220869 by fantastic last updated on 20/May/25 $${Find}\:{the}\:{maximum}\:{value}\:{of}\:{x}^{\mathrm{2}} {y}^{\mathrm{3}} {z}^{\mathrm{4}} \:{subject}\:{to}\:{the}\:{condition}\:{x}+{y}+{z}=\mathrm{18} \\ $$ Commented by mr W last updated on 20/May/25 $${x},\:{y},\:{z}\:\in{N}\:\:\:{or}\:{Z}\:{or}\:{R}\:? \\…
Question Number 220825 by fantastic last updated on 19/May/25 Answered by mr W last updated on 20/May/25 $$\frac{\mathrm{20}}{\mathrm{180}}×\mathrm{72}=\mathrm{8} \\ $$ Commented by mr W last…
Question Number 220810 by Rojarani last updated on 19/May/25 Commented by Ghisom last updated on 19/May/25 $$\mathrm{without}\:\mathrm{further}\:\mathrm{information}\:\mathrm{we}\:\mathrm{can}\:\mathrm{let} \\ $$$${a}={b}={c}={k}\:\Rightarrow \\ $$$${k}=\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{\mathrm{27}}\:\Rightarrow \\ $$$${a}+{b}+{c}=\mathrm{3}{k}=\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{\mathrm{9}}…
Question Number 220811 by aniqah last updated on 19/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com