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Prove-that-there-are-n-r-1-n-1-ways-of-placing-r-identical-objects-in-n-compartments-where-n-gt-r-




Question Number 3884 by Yozzii last updated on 23/Dec/15
Prove that there are  (((n+r−1)),((n−1)) )   ways of placing r identical objects  in n compartments, where n>r.
$${Prove}\:{that}\:{there}\:{are}\:\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}\: \\ $$$${ways}\:{of}\:{placing}\:{r}\:{identical}\:{objects} \\ $$$${in}\:{n}\:{compartments},\:{where}\:{n}>{r}. \\ $$
Commented by RasheedSindhi last updated on 24/Dec/15
Let r objects are denoted as   o_0 ,o_1 ,o_2 ,...o_(r−1) .  For o_0  there are n ways.  For o_1  there are n−1 ways.  For o_2  there are n−2 ways.  For o_3  there are n−3 ways.  ⋮  For o_(r−1)  there are n−r+1 ways.  Total ways:  n(n−1)(n−2)...(n−r+1)  =^n p_r = ((n),(r) ) r!=((n!)/(r!(n−r)!))×r!    =((n!)/((n−r)!))  Or   (((n+r−1)),((n−1)) )  ways.???  Continue
$${Let}\:{r}\:{objects}\:{are}\:{denoted}\:{as}\: \\ $$$$\boldsymbol{\mathrm{o}}_{\mathrm{0}} ,\boldsymbol{\mathrm{o}}_{\mathrm{1}} ,\boldsymbol{\mathrm{o}}_{\mathrm{2}} ,…\boldsymbol{\mathrm{o}}_{\boldsymbol{\mathrm{r}}−\mathrm{1}} . \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{0}} \:{there}\:{are}\:{n}\:{ways}. \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{1}} \:{there}\:{are}\:{n}−\mathrm{1}\:{ways}. \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{2}} \:{there}\:{are}\:{n}−\mathrm{2}\:{ways}. \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\mathrm{3}} \:{there}\:{are}\:{n}−\mathrm{3}\:{ways}. \\ $$$$\vdots \\ $$$${For}\:\boldsymbol{\mathrm{o}}_{\boldsymbol{\mathrm{r}}−\mathrm{1}} \:{there}\:{are}\:{n}−{r}+\mathrm{1}\:{ways}. \\ $$$${Total}\:{ways}: \\ $$$${n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\left({n}−{r}+\mathrm{1}\right) \\ $$$$=^{{n}} \mathrm{p}_{\mathrm{r}} =\begin{pmatrix}{\mathrm{n}}\\{\mathrm{r}}\end{pmatrix}\:\mathrm{r}!=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}−\mathrm{r}\right)!}×\mathrm{r}! \\ $$$$\:\:=\frac{{n}!}{\left({n}−{r}\right)!} \\ $$$${Or} \\ $$$$\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}\:\:{ways}.??? \\ $$$${Continue} \\ $$
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=4 and n is fixed, we could   distribute r=4 like 1+1+1+1 (4 compartments)  or 1+1+2 or 1+2+1 or 2+1+1 (3 compartments)  or 1+3 or 3+1 or 2+2 (2 compartments)  or 4 (1 compartment).  You have n compartments   ⇒  ((n),(4) ) ways for 1+1+1+1  ⇒  ((n),(3) )+ ((n),(3) )+ ((n),(3) )=3 ((n),(3) ) ways   for 1+1+2, 1+2+1, 2+1+1  ⇒3 ((n),(2) ) ways for 1+3,3+1,2+2  ⇒  ((n),(1) ) ways for 4.  ⇒Total number of ways   = ((n),(1) )+3 ((n),(2) )+3 ((n),(3) )+ ((n),(4) ).    For r=5 I deduced   ((n),(1) )+4 ((n),(2) )+6 ((n),(3) )+4 ((n),(4) )+ ((n),(5) ) ways.  I saw the binomial coefficients  arising in these answers so I guessed  that generally,  number of ways=Σ_(i=0) ^(r−1) { (((r−1)),(i) )× ((n),((i+1)) )}.  It is required to prove that    (((n+r−1)),((n−1)) )=Σ_(i=1) ^(r−1) { (((r−1)),(i) ) ((n),((i+1)) )}.
$${From}\:{my}\:{investigation}\:{on}\:{this},\:{you} \\ $$$${can}\:{look}\:{at}\:{how}\:{can}\:{obtain}\:{the}\:{value} \\ $$$${of}\:{r}\:{via}\:{adding}\:{integers}.\:{For}\:{example} \\ $$$${if}\:{r}=\mathrm{4}\:{and}\:{n}\:{is}\:{fixed},\:{we}\:{could}\: \\ $$$${distribute}\:{r}=\mathrm{4}\:{like}\:\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}\:\left(\mathrm{4}\:{compartments}\right) \\ $$$${or}\:\mathrm{1}+\mathrm{1}+\mathrm{2}\:{or}\:\mathrm{1}+\mathrm{2}+\mathrm{1}\:{or}\:\mathrm{2}+\mathrm{1}+\mathrm{1}\:\left(\mathrm{3}\:{compartments}\right) \\ $$$${or}\:\mathrm{1}+\mathrm{3}\:{or}\:\mathrm{3}+\mathrm{1}\:{or}\:\mathrm{2}+\mathrm{2}\:\left(\mathrm{2}\:{compartments}\right) \\ $$$${or}\:\mathrm{4}\:\left(\mathrm{1}\:{compartment}\right). \\ $$$${You}\:{have}\:{n}\:{compartments}\: \\ $$$$\Rightarrow\:\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}\:{ways}\:{for}\:\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$$$\Rightarrow\:\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}=\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}\:{ways}\: \\ $$$${for}\:\mathrm{1}+\mathrm{1}+\mathrm{2},\:\mathrm{1}+\mathrm{2}+\mathrm{1},\:\mathrm{2}+\mathrm{1}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}\:{ways}\:{for}\:\mathrm{1}+\mathrm{3},\mathrm{3}+\mathrm{1},\mathrm{2}+\mathrm{2} \\ $$$$\Rightarrow\:\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\:{ways}\:{for}\:\mathrm{4}. \\ $$$$\Rightarrow{Total}\:{number}\:{of}\:{ways}\: \\ $$$$=\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+\mathrm{3}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}. \\ $$$$ \\ $$$${For}\:{r}=\mathrm{5}\:{I}\:{deduced} \\ $$$$\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+\mathrm{6}\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}+\mathrm{4}\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix}\:{ways}. \\ $$$${I}\:{saw}\:{the}\:{binomial}\:{coefficients} \\ $$$${arising}\:{in}\:{these}\:{answers}\:{so}\:{I}\:{guessed} \\ $$$${that}\:{generally}, \\ $$$${number}\:{of}\:{ways}=\underset{{i}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\sum}}\left\{\begin{pmatrix}{{r}−\mathrm{1}}\\{{i}}\end{pmatrix}×\begin{pmatrix}{{n}}\\{{i}+\mathrm{1}}\end{pmatrix}\right\}. \\ $$$${It}\:{is}\:{required}\:{to}\:{prove}\:{that}\: \\ $$$$\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}=\underset{{i}=\mathrm{1}} {\overset{{r}−\mathrm{1}} {\sum}}\left\{\begin{pmatrix}{{r}−\mathrm{1}}\\{{i}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{i}+\mathrm{1}}\end{pmatrix}\right\}. \\ $$$$ \\ $$$$ \\ $$
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 ′!
$${Perhaps}\:{I}\:{misunderstood}\:{your} \\ $$$${question}.{I}\:{supposed}\:{to}\:{place}\:{one} \\ $$$${object}\:{per}\:{compartment}.{There} \\ $$$${are}\:{n}\:{compartments}.\:{So}\:{first} \\ $$$${of}\:{r}\:{objects}\:{can}\:{go}\:{in}\:{anyone} \\ $$$${of}\:{n}\:{compartments}.{So}\:{there}\:{are} \\ $$$${are}\:{n}\:{ways}\:{for}\:{first}\:{object}.\:{Actually} \\ $$$${I}\:{didn}'{t}\:{give}\:{due}\:{importance}\:{to} \\ $$$${the}\:{word}\:'\:{identical}\:'! \\ $$
Commented by prakash jain last updated on 24/Dec/15
Stars and Bars
$$\mathrm{Stars}\:\mathrm{and}\:\mathrm{Bars} \\ $$
Answered by prakash jain last updated on 24/Dec/15
Taks (n+r−1) position.   Place n−1 markers at any of those postions.  Remaining r positions are occupied by  objects.  (n−1) markers divide r objects in n−compartments.  So total number of ways: ^(n+r−1) C_(n−1) .  Examples (stars are objects bars divide them)  4 obects 6 compartments (5 bars)  ★∣∣★∣∣∣★★    divisions 1,0,1,0,0,2  total position 9  − − − − − − − − −  place (n−1) bars or dividers  − ∣  ∣ −∣ ∣ ∣ − −  r objects arr divided into n−compartments.  This method is known as stars and bars.
$$\mathrm{Taks}\:\left({n}+{r}−\mathrm{1}\right)\:\mathrm{position}.\: \\ $$$$\mathrm{Place}\:{n}−\mathrm{1}\:\mathrm{markers}\:\mathrm{at}\:\mathrm{any}\:\mathrm{of}\:\mathrm{those}\:\mathrm{postions}. \\ $$$$\mathrm{Remaining}\:{r}\:\mathrm{positions}\:\mathrm{are}\:\mathrm{occupied}\:\mathrm{by} \\ $$$$\mathrm{objects}. \\ $$$$\left({n}−\mathrm{1}\right)\:{markers}\:{divide}\:{r}\:{objects}\:{in}\:{n}−{compartments}. \\ $$$${S}\mathrm{o}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}:\:\:^{{n}+{r}−\mathrm{1}} {C}_{{n}−\mathrm{1}} . \\ $$$$\mathrm{Examples}\:\left(\mathrm{stars}\:\mathrm{are}\:\mathrm{objects}\:\mathrm{bars}\:\mathrm{divide}\:\mathrm{them}\right) \\ $$$$\mathrm{4}\:\mathrm{obects}\:\mathrm{6}\:\mathrm{compartments}\:\left(\mathrm{5}\:\mathrm{bars}\right) \\ $$$$\bigstar\mid\mid\bigstar\mid\mid\mid\bigstar\bigstar\: \\ $$$$\:\mathrm{divisions}\:\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{0},\mathrm{0},\mathrm{2} \\ $$$$\mathrm{total}\:\mathrm{position}\:\mathrm{9} \\ $$$$−\:−\:−\:−\:−\:−\:−\:−\:− \\ $$$$\mathrm{place}\:\left({n}−\mathrm{1}\right)\:\mathrm{bars}\:\mathrm{or}\:\mathrm{dividers} \\ $$$$−\:\mid\:\:\mid\:−\mid\:\mid\:\mid\:−\:− \\ $$$${r}\:{objects}\:{arr}\:{divided}\:{into}\:{n}−{compartments}. \\ $$$$\mathrm{This}\:\mathrm{method}\:\mathrm{is}\:\mathrm{known}\:\mathrm{as}\:\mathrm{stars}\:\mathrm{and}\:\mathrm{bars}. \\ $$
Commented by Yozzii last updated on 24/Dec/15
I never heard of this method before.  I went after considering the identity   (1+x)^(n+r−1) =(1+x)^n (1+x)^(r−1)    for the coefficient in x^(n−1)   to prove that Σ_(i=0) ^(r−1) { (((r−1)),(i) ) ((n),((i+1)) )}= (((n+r−1)),((n−1)) ).
$${I}\:{never}\:{heard}\:{of}\:{this}\:{method}\:{before}. \\ $$$${I}\:{went}\:{after}\:{considering}\:{the}\:{identity} \\ $$$$\:\left(\mathrm{1}+{x}\right)^{{n}+{r}−\mathrm{1}} =\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{r}−\mathrm{1}} \: \\ $$$${for}\:{the}\:{coefficient}\:{in}\:{x}^{{n}−\mathrm{1}} \\ $$$${to}\:{prove}\:{that}\:\underset{{i}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\sum}}\left\{\begin{pmatrix}{{r}−\mathrm{1}}\\{{i}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{i}+\mathrm{1}}\end{pmatrix}\right\}=\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{{n}−\mathrm{1}}\end{pmatrix}. \\ $$$$ \\ $$
Commented by prakash jain last updated on 24/Dec/15
I think it is truly exceptional on your  part to come up with this method on your  own.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{truly}\:\mathrm{exceptional}\:\mathrm{on}\:\mathrm{your} \\ $$$$\mathrm{part}\:\mathrm{to}\:\mathrm{come}\:\mathrm{up}\:\mathrm{with}\:\mathrm{this}\:\mathrm{method}\:\mathrm{on}\:\mathrm{your} \\ $$$$\mathrm{own}.\: \\ $$
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.
$${Thank}\:{you}\:{very}\:{much}\:{for}\:{your}\:{kind} \\ $$$${words}!\:{Encouragement}\:{is}\:{beneficial} \\ $$$${when}\:{it}\:{is}\:{you}\:{think}\:{you}'{re}\:{making} \\ $$$${slow}\:{progress}\:{in}\:{individual}\:{mathematical} \\ $$$${studies}.\:{You}'{re}\:{further}\:{driven}\:{to} \\ $$$${continue}\:{trying}\:{as}\:{such}\:{a}\:{person}. \\ $$

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