Question Number 3884 by Yozzii last updated on 23/Dec/15
$${Prove}\:{that}\:{there}\:{are}\:\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}\: \\ $$$${ways}\:{of}\:{placing}\:{r}\:{identical}\:{objects} \\ $$$${in}\:{n}\:{compartments},\:{where}\:{n}>{r}. \\ $$
Commented by RasheedSindhi last updated on 24/Dec/15
$${Let}\:{r}\:{objects}\:{are}\:{denoted}\:{as}\: \\ $$$$\boldsymbol{\mathrm{o}}_{\mathrm{0}} ,\boldsymbol{\mathrm{o}}_{\mathrm{1}} ,\boldsymbol{\mathrm{o}}_{\mathrm{2}} ,…\boldsymbol{\mathrm{o}}_{\boldsymbol{\mathrm{r}}−\mathrm{1}} . \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{0}} \:{there}\:{are}\:{n}\:{ways}. \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{1}} \:{there}\:{are}\:{n}−\mathrm{1}\:{ways}. \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{2}} \:{there}\:{are}\:{n}−\mathrm{2}\:{ways}. \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{3}} \:{there}\:{are}\:{n}−\mathrm{3}\:{ways}. \\ $$$$\vdots \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\boldsymbol{\mathrm{r}}−\mathrm{1}} \:{there}\:{are}\:{n}−{r}+\mathrm{1}\:{ways}. \\ $$$${Total}\:{ways}: \\ $$$${n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\left({n}−{r}+\mathrm{1}\right) \\ $$$$=^{{n}} \mathrm{p}_{\mathrm{r}} =\begin{pmatrix}{\mathrm{n}}\\{\mathrm{r}}\end{pmatrix}\:\mathrm{r}!=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}−\mathrm{r}\right)!}×\mathrm{r}! \\ $$$$\:\:=\frac{{n}!}{\left({n}−{r}\right)!} \\ $$$${Or} \\ $$$$\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}\:\:{ways}.??? \\ $$$${Continue} \\ $$
Commented by Yozzii last updated on 24/Dec/15
$${From}\:{my}\:{investigation}\:{on}\:{this},\:{you} \\ $$$${can}\:{look}\:{at}\:{how}\:{can}\:{obtain}\:{the}\:{value} \\ $$$${of}\:{r}\:{via}\:{adding}\:{integers}.\:{For}\:{example} \\ $$$${if}\:{r}=\mathrm{4}\:{and}\:{n}\:{is}\:{fixed},\:{we}\:{could}\: \\ $$$${distribute}\:{r}=\mathrm{4}\:{like}\:\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}\:\left(\mathrm{4}\:{compartments}\right) \\ $$$${or}\:\mathrm{1}+\mathrm{1}+\mathrm{2}\:{or}\:\mathrm{1}+\mathrm{2}+\mathrm{1}\:{or}\:\mathrm{2}+\mathrm{1}+\mathrm{1}\:\left(\mathrm{3}\:{compartments}\right) \\ $$$${or}\:\mathrm{1}+\mathrm{3}\:{or}\:\mathrm{3}+\mathrm{1}\:{or}\:\mathrm{2}+\mathrm{2}\:\left(\mathrm{2}\:{compartments}\right) \\ $$$${or}\:\mathrm{4}\:\left(\mathrm{1}\:{compartment}\right). \\ $$$${You}\:{have}\:{n}\:{compartments}\: \\ $$$$\Rightarrow\:\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}\:{ways}\:{for}\:\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$\Rightarrow\:\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}=\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:{ways}\: \\ $$$${for}\:\mathrm{1}+\mathrm{1}+\mathrm{2},\:\mathrm{1}+\mathrm{2}+\mathrm{1},\:\mathrm{2}+\mathrm{1}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:{ways}\:{for}\:\mathrm{1}+\mathrm{3},\mathrm{3}+\mathrm{1},\mathrm{2}+\mathrm{2} \\ $$$$\Rightarrow\:\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:{ways}\:{for}\:\mathrm{4}. \\ $$$$\Rightarrow{Total}\:{number}\:{of}\:{ways}\: \\ $$$$=\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}. \\ $$$$ \\ $$$${For}\:{r}=\mathrm{5}\:{I}\:{deduced} \\ $$$$\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+\mathrm{6}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix}\:{ways}. \\ $$$${I}\:{saw}\:{the}\:{binomial}\:{coefficients} \\ $$$${arising}\:{in}\:{these}\:{answers}\:{so}\:{I}\:{guessed} \\ $$$${that}\:{generally}, \\ $$$${number}\:{of}\:{ways}=\underset{{i}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\sum}}\left\{\begin{pmatrix}{{r}−\mathrm{1}}\\{{i}}\end{pmatrix}×\begin{pmatrix}{{n}}\\{{i}+\mathrm{1}}\end{pmatrix}\right\}. \\ $$$${It}\:{is}\:{required}\:{to}\:{prove}\:{that}\: \\ $$$$\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}=\underset{{i}=\mathrm{1}} {\overset{{r}−\mathrm{1}} {\sum}}\left\{\begin{pmatrix}{{r}−\mathrm{1}}\\{{i}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{i}+\mathrm{1}}\end{pmatrix}\right\}. \\ $$$$ \\ $$$$ \\ $$
Commented by RasheedSindhi last updated on 24/Dec/15
$${Perhaps}\:{I}\:{misunderstood}\:{your} \\ $$$${question}.{I}\:{supposed}\:{to}\:{place}\:{one} \\ $$$${object}\:{per}\:{compartment}.{There} \\ $$$${are}\:{n}\:{compartments}.\:{So}\:{first} \\ $$$${of}\:{r}\:{objects}\:{can}\:{go}\:{in}\:{anyone} \\ $$$${of}\:{n}\:{compartments}.{So}\:{there}\:{are} \\ $$$${are}\:{n}\:{ways}\:{for}\:{first}\:{object}.\:{Actually} \\ $$$${I}\:{didn}'{t}\:{give}\:{due}\:{importance}\:{to} \\ $$$${the}\:{word}\:'\:{identical}\:'! \\ $$
Commented by prakash jain last updated on 24/Dec/15
$$\mathrm{Stars}\:\mathrm{and}\:\mathrm{Bars} \\ $$
Answered by prakash jain last updated on 24/Dec/15
$$\mathrm{Taks}\:\left({n}+{r}−\mathrm{1}\right)\:\mathrm{position}.\: \\ $$$$\mathrm{Place}\:{n}−\mathrm{1}\:\mathrm{markers}\:\mathrm{at}\:\mathrm{any}\:\mathrm{of}\:\mathrm{those}\:\mathrm{postions}. \\ $$$$\mathrm{Remaining}\:{r}\:\mathrm{positions}\:\mathrm{are}\:\mathrm{occupied}\:\mathrm{by} \\ $$$$\mathrm{objects}. \\ $$$$\left({n}−\mathrm{1}\right)\:{markers}\:{divide}\:{r}\:{objects}\:{in}\:{n}−{compartments}. \\ $$$${S}\mathrm{o}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}:\:\:^{{n}+{r}−\mathrm{1}} {C}_{{n}−\mathrm{1}} . \\ $$$$\mathrm{Examples}\:\left(\mathrm{stars}\:\mathrm{are}\:\mathrm{objects}\:\mathrm{bars}\:\mathrm{divide}\:\mathrm{them}\right) \\ $$$$\mathrm{4}\:\mathrm{obects}\:\mathrm{6}\:\mathrm{compartments}\:\left(\mathrm{5}\:\mathrm{bars}\right) \\ $$$$\bigstar\mid\mid\bigstar\mid\mid\mid\bigstar\bigstar\: \\ $$$$\:\mathrm{divisions}\:\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{0},\mathrm{0},\mathrm{2} \\ $$$$\mathrm{total}\:\mathrm{position}\:\mathrm{9} \\ $$$$−\:−\:−\:−\:−\:−\:−\:−\:− \\ $$$$\mathrm{place}\:\left({n}−\mathrm{1}\right)\:\mathrm{bars}\:\mathrm{or}\:\mathrm{dividers} \\ $$$$−\:\mid\:\:\mid\:−\mid\:\mid\:\mid\:−\:− \\ $$$${r}\:{objects}\:{arr}\:{divided}\:{into}\:{n}−{compartments}. \\ $$$$\mathrm{This}\:\mathrm{method}\:\mathrm{is}\:\mathrm{known}\:\mathrm{as}\:\mathrm{stars}\:\mathrm{and}\:\mathrm{bars}. \\ $$
Commented by Yozzii last updated on 24/Dec/15
$${I}\:{never}\:{heard}\:{of}\:{this}\:{method}\:{before}. \\ $$$${I}\:{went}\:{after}\:{considering}\:{the}\:{identity} \\ $$$$\:\left(\mathrm{1}+{x}\right)^{{n}+{r}−\mathrm{1}} =\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{r}−\mathrm{1}} \: \\ $$$${for}\:{the}\:{coefficient}\:{in}\:{x}^{{n}−\mathrm{1}} \\ $$$${to}\:{prove}\:{that}\:\underset{{i}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\sum}}\left\{\begin{pmatrix}{{r}−\mathrm{1}}\\{{i}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{i}+\mathrm{1}}\end{pmatrix}\right\}=\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}. \\ $$$$ \\ $$
Commented by prakash jain last updated on 24/Dec/15
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{truly}\:\mathrm{exceptional}\:\mathrm{on}\:\mathrm{your} \\ $$$$\mathrm{part}\:\mathrm{to}\:\mathrm{come}\:\mathrm{up}\:\mathrm{with}\:\mathrm{this}\:\mathrm{method}\:\mathrm{on}\:\mathrm{your} \\ $$$$\mathrm{own}.\: \\ $$
Commented by Yozzii last updated on 24/Dec/15
$${Thank}\:{you}\:{very}\:{much}\:{for}\:{your}\:{kind} \\ $$$${words}!\:{Encouragement}\:{is}\:{beneficial} \\ $$$${when}\:{it}\:{is}\:{you}\:{think}\:{you}'{re}\:{making} \\ $$$${slow}\:{progress}\:{in}\:{individual}\:{mathematical} \\ $$$${studies}.\:{You}'{re}\:{further}\:{driven}\:{to} \\ $$$${continue}\:{trying}\:{as}\:{such}\:{a}\:{person}. \\ $$