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Number-of-distributions-of-n-different-articles-to-r-different-boxes-so-as-1-empty-box-allowed-2-empty-box-not-allowed-with-proof-kindly-help-me-

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Question Number 195538 by SLVR last updated on 04/Aug/23
Number of distributions of n different articles to r different boxes so as 1)empty box allowed 2)empty box not allowed with proof...kindly help me
$${Number}\:{of}\:{distributions}\:{of} \\ $$$${n}\:{different}\:{articles}\:{to}\:{r}\:{different}\:\:{boxes} \\ $$$$\left.{so}\:{as}\:\mathrm{1}\right){empty}\:{box}\:{allowed} \\ $$$$\left.\mathrm{2}\right){empty}\:{box}\:{not}\:{allowed} \\ $$$${with}\:{proof}…{kindly}\:{help}\:{me} \\ $$
Commented by SLVR last updated on 04/Aug/23
sir...kindly give proof
$${sir}…{kindly}\:{give}\:{proof} \\ $$
Commented by SLVR last updated on 05/Aug/23
sir Mr.W..or...any one can help me...i need bit urgently
$${sir}\:{Mr}.{W}..{or}…{any}\:{one}\:{can} \\ $$$${help}\:{me}…{i}\:{need}\:{bit}\:{urgently} \\ $$
Commented by mr W last updated on 05/Aug/23
using generating function method we can see that the result is the coefficient of term x^n in expansion 1) n!(1+(x/1)+(x^2 /(2!))+(x^3 /(3!))+...)^r , i.e. n!e^(rx) 2) n!((x/1)+(x^2 /(2!))+(x^3 /(3!))+...)^r , i.e. n!(e^x −1)^r
$${using}\:{generating}\:{function}\:{method} \\ $$$${we}\:{can}\:{see}\:{that}\:{the}\:{result}\:{is}\:{the} \\ $$$${coefficient}\:{of}\:{term}\:{x}^{{n}} \:{in}\:{expansion} \\ $$$$\left.\mathrm{1}\right)\:{n}!\left(\mathrm{1}+\frac{{x}}{\mathrm{1}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…\right)^{{r}} ,\:{i}.{e}.\:\:{n}!{e}^{{rx}} \\ $$$$\left.\mathrm{2}\right)\:{n}!\left(\frac{{x}}{\mathrm{1}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…\right)^{{r}} ,\:{i}.{e}.\:\:{n}!\left({e}^{{x}} −\mathrm{1}\right)^{{r}} \\ $$
Commented by SLVR last updated on 05/Aug/23
Thanks Prof.W...so kind of you
$${Thanks}\:{Prof}.{W}…{so}\:{kind}\:{of}\:{you} \\ $$
Commented by mr W last updated on 05/Aug/23
a little explanation: say box 1 has k_1 objects, box 2 has k_2 objects, etc. with k_1 +k_2 +...k_r =n and k_1 ,k_2 ,...,k_r are fixed values. to distribute n objects in r boxes in this way there are ((n!)/((k_1 )!×(k_2 )!×...(k_r )!)) possibilities. since k_1 ,k_2 ,...,k_r can be variable values, we can construct a generating function like n!((x/(1!))+(x^2 /(2!))+...+...)^r if each box must get at least an object, or n!(1+(x/(1!))+(x^2 /(2!))+...+...)^r if each box may also be empty. for 1) the formula is r^n . for 2) the formula is r^n −C_1 ^r (r−1)^n +C_2 ^r (r−2)^n −...+(−1)^(n−1) C_(r−1) ^r
$${a}\:{little}\:{explanation}: \\ $$$${say}\:{box}\:\mathrm{1}\:{has}\:{k}_{\mathrm{1}} \:{objects},\:{box}\:\mathrm{2}\:{has}\:{k}_{\mathrm{2}} \\ $$$${objects},\:{etc}.\:{with}\:{k}_{\mathrm{1}} +{k}_{\mathrm{2}} +…{k}_{{r}} ={n}\:{and} \\ $$$${k}_{\mathrm{1}} ,{k}_{\mathrm{2}} ,…,{k}_{{r}} \:{are}\:{fixed}\:{values}. \\ $$$${to}\:{distribute}\:{n}\:{objects}\:{in}\:{r}\:{boxes}\:{in} \\ $$$${this}\:{way}\:{there}\:{are}\:\frac{{n}!}{\left({k}_{\mathrm{1}} \right)!×\left({k}_{\mathrm{2}} \right)!×…\left({k}_{{r}} \right)!} \\ $$$${possibilities}. \\ $$$${since}\:{k}_{\mathrm{1}} ,{k}_{\mathrm{2}} ,…,{k}_{{r}} \:{can}\:{be}\:{variable}\:{values}, \\ $$$${we}\:{can}\:{construct}\:{a}\:{generating} \\ $$$${function}\:{like} \\ $$$${n}!\left(\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+…+…\right)^{{r}} \:{if}\:{each}\:{box}\:{must} \\ $$$${get}\:{at}\:{least}\:{an}\:{object},\:{or} \\ $$$${n}!\left(\mathrm{1}+\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+…+…\right)^{{r}} \:{if}\:{each}\:{box}\:{may} \\ $$$${also}\:{be}\:{empty}. \\ $$$$ \\ $$$$\left.{for}\:\mathrm{1}\right)\:{the}\:{formula}\:{is}\:{r}^{{n}} . \\ $$$$\left.{for}\:\mathrm{2}\right)\:{the}\:{formula}\:{is}\: \\ $$$$\:\:\:{r}^{{n}} −{C}_{\mathrm{1}} ^{{r}} \left({r}−\mathrm{1}\right)^{{n}} +{C}_{\mathrm{2}} ^{{r}} \left({r}−\mathrm{2}\right)^{{n}} −…+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {C}_{{r}−\mathrm{1}} ^{{r}} \\ $$
Commented by mr W last updated on 05/Aug/23
example: we have 10 foreign students who should be distributed to 3 schools. in how many ways can this be done? 1) 3^(10) =59049 ways 2) 3^(10) −3×2^(10) +3×1^(10) =55980 ways
$${example}: \\ $$$${we}\:{have}\:\mathrm{10}\:{foreign}\:{students}\:{who} \\ $$$${should}\:{be}\:{distributed}\:{to}\:\mathrm{3}\:{schools}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{this}\:{be}\:{done}? \\ $$$$\left.\mathrm{1}\right)\:\mathrm{3}^{\mathrm{10}} =\mathrm{59049}\:{ways} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{3}^{\mathrm{10}} −\mathrm{3}×\mathrm{2}^{\mathrm{10}} +\mathrm{3}×\mathrm{1}^{\mathrm{10}} =\mathrm{55980}\:{ways} \\ $$
Commented by mr W last updated on 05/Aug/23
Answered by MM42 last updated on 05/Aug/23
if n≥r let x_i = number articles in the i^(th) box 1)⇒ x_1 +x_2 +...+x_r =n ; x_i ≥0 ans= (((n+r−1)),(( n)) )=(((n+r−1)!)/(n!(r−1)!)) 2)⇒ x_1 +x_2 +...+x_r =n ; x_i ≥1 ⇒ y_1 +y_2 +...+y_r =n−r ; y_i ≥0 ans= (((n−1)),((n−r)) )=(((n−1)!)/((n−r)!(r−1)!)) if n<r then no all box can be full.
$${if}\:\:{n}\geqslant{r} \\ $$$${let}\:\:{x}_{{i}} \:=\:{number}\:{articles}\:{in}\:{the}\:{i}^{{th}} \:{box} \\ $$$$\left.\mathrm{1}\right)\Rightarrow\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +…+{x}_{{r}} ={n}\:\:;\:\:{x}_{{i}} \:\geqslant\mathrm{0} \\ $$$${ans}=\begin{pmatrix}{{n}+{r}−\mathrm{1}}\\{\:\:\:\:\:\:\:\:\:{n}}\end{pmatrix}=\frac{\left({n}+{r}−\mathrm{1}\right)!}{{n}!\left({r}−\mathrm{1}\right)!} \\ $$$$\left.\mathrm{2}\right)\Rightarrow\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +…+{x}_{{r}} ={n}\:\:;\:\:{x}_{{i}} \:\geqslant\mathrm{1} \\ $$$$\Rightarrow\:{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +…+{y}_{{r}} ={n}−{r}\:\:;\:\:{y}_{{i}} \:\geqslant\mathrm{0} \\ $$$${ans}=\begin{pmatrix}{{n}−\mathrm{1}}\\{{n}−{r}}\end{pmatrix}=\frac{\left({n}−\mathrm{1}\right)!}{\left({n}−{r}\right)!\left({r}−\mathrm{1}\right)!} \\ $$$${if}\:\:{n}<{r}\:\:{then}\:\:{no}\:{all}\:{box}\:{can}\:{be}\:{full}. \\ $$$$ \\ $$
Commented by mr W last updated on 05/Aug/23
what you solved is for the case that the objects are identical. but the question said that both the boxes and the objects are different.
$${what}\:{you}\:{solved}\:{is}\:{for}\:{the}\:{case}\:{that} \\ $$$${the}\:{objects}\:{are}\:{identical}.\:{but}\:{the} \\ $$$${question}\:{said}\:{that}\:{both}\:{the}\:{boxes} \\ $$$${and}\:{the}\:{objects}\:{are}\:{different}. \\ $$
Commented by MM42 last updated on 08/Aug/23
yes.you are right.i did not notice the diffrence between the subjects. thanks for you
$${yes}.{you}\:{are}\:{right}.{i}\:{did}\:{not}\:{notice} \\ $$$${the}\:{diffrence}\:{between}\:{the}\:{subjects}. \\ $$$${thanks}\:{for}\:{you}\: \\ $$$$ \\ $$

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