Question Number 207374 by sniper237 last updated on 13/May/24
$${Show}\:{that}\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$
Answered by mr W last updated on 13/May/24
$$\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} =\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \\ $$$$\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {x}^{{k}} \right)\left(\underset{{s}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{s}} ^{{n}} {x}^{{s}} \right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{C}_{{r}} ^{\mathrm{2}{n}} {x}^{{r}} \\ $$$${coef}.\:{of}\:{x}^{{n}} : \\ $$$${k}+{s}={n},\:{r}={n} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {C}_{{s}} ^{{n}} ={C}_{{n}} ^{\mathrm{2}{n}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {C}_{{n}−{k}} ^{{n}} ={C}_{{n}} ^{\mathrm{2}{n}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} ={C}_{{n}} ^{\mathrm{2}{n}} \\ $$
Answered by mr W last updated on 13/May/24
$${an}\:{other}\:{method} \\ $$$${say}\:{there}\:{are}\:{n}\:{men}\:{and}\:{n}\:{women}. \\ $$$${to}\:{select}\:{n}\:{persons}\:{from}\:{them}\:{there} \\ $$$${are}\:{C}_{{n}} ^{\mathrm{2}{n}} \:{ways}.\:\:{we}\:{can}\:{also}\:{select} \\ $$$${k}\:{men}\:{and}\:{n}−{k}\:{women},\:{there}\:{are} \\ $$$${C}_{{k}} ^{{n}} {C}_{{n}−{k}} ^{{n}} =\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} \:{ways}.\:{k}\:{can}\:{be}\:{from} \\ $$$$\mathrm{0}\:{to}\:{n},\:{therefore}\:{totally}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} \:{ways}. \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} ={C}_{{n}} ^{\mathrm{2}{n}} \\ $$