Question Number 224261 by Tawa11 last updated on 30/Aug/25 If I have 10 people and I want to create pairings, so that every person is…
Question Number 222192 by mr W last updated on 22/Jun/25 $${there}\:{are}\:\mathrm{32}\:{students}\:{in}\:{a}\:{class}.\:{for} \\ $$$${each}\:{competition}\:{in}\:{a}\:{sport}\:{event}\: \\ $$$${in}\:{the}\:{school}\:{each}\:{class}\:{can}\:{send} \\ $$$${a}\:{team}\:{with}\:{three}\:{students}.\:{if}\:{no} \\ $$$${two}\:{students}\:{may}\:{be}\:{in}\:{the}\:{same} \\ $$$${team}\:{for}\:{more}\:{than}\:{one}\:{time},\:{in} \\ $$$${how}\:{many}\:{different}\:{competitions}\: \\ $$$${can}\:{this}\:{class}\:{participate}?…
Question Number 222105 by alcohol last updated on 17/Jun/25 Commented by alcohol last updated on 17/Jun/25 $${please}\:{help}\:{with}\:\left({iii}\right)\:{a}\:{and}\:{b}\: \\ $$ Answered by mr W last updated…
Question Number 221661 by mr W last updated on 09/Jun/25 Commented by mr W last updated on 09/Jun/25 $${an}\:{unsolved}\:{old}\:{question} \\ $$ Answered by MrGaster last…
Question Number 220877 by Spillover last updated on 20/May/25 Answered by Rasheed.Sindhi last updated on 20/May/25 $$\left({a}\right)\:\left({n}+\mathrm{2}\right)!+\left({n}+\mathrm{1}\right)!+{n}! \\ $$$$=\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}!+\left({n}+\mathrm{1}\right){n}!+{n}! \\ $$$$={n}!\left(\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)+\left({n}+\mathrm{1}\right)+\mathrm{1}\right) \\ $$$$\left.={n}!\left\{\:\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)+\mathrm{1}\right)\right\} \\ $$$$={n}!\left({n}^{\mathrm{2}}…
Question Number 220876 by Spillover last updated on 20/May/25 Answered by Rasheed.Sindhi last updated on 20/May/25 $$\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 220878 by Spillover last updated on 20/May/25 Answered by Rasheed.Sindhi last updated on 22/May/25 $$\frac{{n}!}{\left({n}−{r}\right)!{r}!}+\frac{\mathrm{2}×{n}!}{\left({n}−{r}+\mathrm{1}\right)!\left({r}−\mathrm{1}\right)!}+\frac{{n}!}{\left({n}−{r}+\mathrm{2}\right)!\left({r}−\mathrm{2}\right)!} \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!{r}\left({r}−\mathrm{1}\right)\left({r}−\mathrm{2}\right)!}+\frac{\mathrm{2}×{n}!}{\left({n}−{r}+\mathrm{1}\right)\left({n}−{r}\right)!\left({r}−\mathrm{1}\right)\left({r}−\mathrm{2}\right)!}+\frac{{n}!}{\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!} \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!}\left(\frac{\mathrm{1}}{{r}\left({r}−\mathrm{1}\right)}+\frac{\mathrm{2}}{\left({n}−{r}+\mathrm{1}\right)\left({r}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)}\right) \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!}\left(\frac{\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)+\mathrm{2}{r}\left({n}−{r}+\mathrm{2}\right)+{r}\left({r}−\mathrm{1}\right)}{{r}\left({r}−\mathrm{1}\right)\left({n}−{r}+\mathrm{2}\right)\left({n}−{r}+\mathrm{1}\right)}\right) \\ $$$$\frac{{n}!}{\left({n}−{r}\right)!\left({r}−\mathrm{2}\right)!}\left(\frac{\left({n}−{r}\right)^{\mathrm{2}}…
Question Number 220873 by Spillover last updated on 20/May/25 Answered by mr W last updated on 20/May/25 $$\mathrm{18}!×\mathrm{19}×\mathrm{18}=\mathrm{18}×\mathrm{19}! \\ $$$${or} \\ $$$$\mathrm{20}!−\mathrm{2}×\mathrm{19}!=\mathrm{18}×\mathrm{19}! \\ $$ Commented…
Question Number 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 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} \\…