Question Number 160493 by SANOGO last updated on 30/Nov/21
$${montrer}\:{a}\:{l}\:{aide}\:{de}\:{binome}\:{de}\:{newton}\:{que}:\: \\ $$$$\underset{{k}={o}} {\overset{{r}} {\sum}}\left(\underset{{k}} {\:}^{{n}} \right)\left(_{{r}−{k}} ^{{m}} \right)=\left(_{\:\:\:\:\:{r}} ^{{m}+{n}} \right)\: \\ $$
Answered by mr W last updated on 30/Nov/21
$${say}\:{there}\:{are}\:{m}\:{men}\:{and}\:{n}\:{women}\:{in} \\ $$$${a}\:{conference}.\:{r}\:{persons}\:{should}\:{be}\: \\ $$$${selected}\:{to}\:{build}\:{a}\:{committee}.\:{in}\:{how} \\ $$$${many}\:{ways}\:{can}\:{this}\:{be}\:{done}? \\ $$$$\mathrm{0}\:{woman},\:{r}\:{men}:\:\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{{m}}\\{{r}}\end{pmatrix}\:{ways} \\ $$$$\mathrm{1}\:{woman},\:{r}−\mathrm{1}\:{men}:\:\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{m}}\\{{r}−\mathrm{1}}\end{pmatrix}\:{ways} \\ $$$$\mathrm{2}\:{women},\:{r}−\mathrm{2}\:{men}:\:\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\begin{pmatrix}{{m}}\\{{r}−\mathrm{2}}\end{pmatrix}\:{ways} \\ $$$${k}\:{women},\:{r}−{k}\:{men}:\:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\begin{pmatrix}{{m}}\\{{r}−{k}}\end{pmatrix}\:{ways} \\ $$$$…. \\ $$$${r}\:{women},\:\mathrm{0}\:{man}:\:\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\begin{pmatrix}{{m}}\\{\mathrm{0}}\end{pmatrix}\:{ways} \\ $$$${totally}\:\underset{{k}=\mathrm{0}} {\overset{{r}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\begin{pmatrix}{{m}}\\{{r}−{k}}\end{pmatrix}\:{ways}. \\ $$$${on}\:{the}\:{other}\:{side}\:{to}\:{select}\:{r}\:{persons} \\ $$$${from}\:{n}+{m}\:{persons},\:{there}\:{are} \\ $$$$\begin{pmatrix}{{m}+{n}}\\{{r}}\end{pmatrix}\:{ways},\:{therefore} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{r}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\begin{pmatrix}{{m}}\\{{r}−{k}}\end{pmatrix}\:=\begin{pmatrix}{{m}+{n}}\\{{r}}\end{pmatrix} \\ $$