 the divider.
As a side note, we've only shown that $\Omega_\text{sys}$ and $\Omega_\text{sys}'$ are equal when ignoring low order terms. Actually, we should not expect them to be equal exactly either. Before removing the divider, each box contained a definite number of particles, but after the divider is removed, the particle number in each box will be variable.
Thus, the Gibbs paradox is resolved without needing to use correct Boltzmann counting. If the "naive" approach to calculating the multiplicity was correct all along, why should we still insist on dividing by $N!$ to account for indistinguishability? How is division by $N!$ even justified?
There is one downside to not using correct Boltzmann counting, namely, entropy is no longer extensive. By an extensive entropy, I mean, one where the following is true. The entropy of a total system, is equal to the sum of the entropies of 'each subsystem when considered in isolation', regardless if some subsystems are "identical". This is equivalent to requiring that the multiplicity of the total system, be a product of the multiplicities of 'each subsystem when considered in isolation'.
$$
\Omega_\text{total} = \prod_i \Omega_\text{subsystem $i$}
$$
The multiplicity we have been using so far, clearly does not satisfy this property, we had to multiply by a multinomial coefficient to get the correct total system multiplicity;
$$
\Omega_\text{total} \neq \Omega_\text{left box}\times\Omega_\text{right box} \,,
$$
where $\Omega_\text{left box}$ is the multiplicity of the left box, when the left box is considered in isolation, and similarly for $\Omega_\text{right box}$.
There is a natural connection between how we have been calculating multiplicities so far, and correct Boltzmann counting. By rearranging the equation relating the total system multiplicity (before the divider was removed) and the multiplicities of the left and right boxes (when considered in isolation), we find
\begin{gather*}
\Omega_\text{sys}
    = \Omega(V_1, N_1)\Omega(V_2, N_2) \times  \frac{(N_1+N_2)!}{N_1!N_2!} \\
\frac{\Omega_\text{sys}}{(N_
the divider. 
 As a side note, we've only testify that $ \Omega_\text{sys}$ and $ \Omega_\text{sys}'$ are adequate when ignoring low decree price. Actually, we should not expect them to be equal precisely either. Before removing the divider, each box incorporate a definite number of particles, but after the partition is removed, the particle phone number in each corner will be variable. 
 Thus, the Gibbs paradox is purpose without needing to use right Boltzmann counting. If the " naive " approach to calculate the multiplicity was correct wholly along, why should we still importune on dividing by $ N!$ to account for indistinguishability? How is division by $ N!$ even justified? 
 There is one downside to not using right Boltzmann counting, namely, entropy is no longer extensive. By an extensive entropy, I mean, one where the following is genuine. The entropy of a total system, is equal to the sum of the entropies of' each subsystem when considered in isolation', regardless if some subsystems are " identical ". This is equivalent to requiring that the multiplicity of the entire system, be a product of the numerousness of' each subsystem when consider in isolation'. 
 $ $ 
 \Omega_\text{total } = \prod_i \Omega_\text{subsystem $ i$ } 
 $ $ 
 The multiplicity we have been using so far, intelligibly does not satisfy this property, we had to multiply by a polynomial coefficient to get the correct total system multiplicity; 
 $ $ 
 \Omega_\text{total } \neq \Omega_\text{left box}\times\Omega_\text{right box } \, , 
 $ $ 
 where $ \Omega_\text{left box}$ is the multiplicity of the left box, when the left box is considered in isolation, and similarly for $ \Omega_\text{right box}$. 
 There is a natural connection between how we have been calculating multiplicity so far, and decline Boltzmann counting. By rearranging the equation relating the total arrangement multiplicity (before the divider was removed) and the multiplicity of the left and right boxes (when considered in isolation), we line up 
 \begin{gather * } 
 \Omega_\text{sys } 
     = \Omega(V_1, N_1)\Omega(V_2, N_2) \times   \frac{(N_1+N_2)!}{N_1!N_2! } \\ 
 \frac{\Omega_\text{sys}}{(N _
 thf divider.
As a side note, de've only shown that $\Okega_\test{sys}$ ana $\Omega_\text{sys}'$ are equal whei igborint low order terms. Actuxlly, we sjould nor exkect them to be equal exagcly ejbher. Yehore removing tme divider, aach box contahndd a definite number of particles, but after yhf divider is rgmovec, the iavticle number in each box will bs variaule.
Thus, the Gibns paradox is resolved witjout needing to use cogrect Boltznann xounting. If ghe "naive" approach to dalculating the multiplicity war corxect all alint, wji should we wtill insist on dlniding ty $N!$ to account for ikdistmnguushability? How is divmsion by $N!$ even justyfied?
Thera ns one downside to nor ysing corsect Volgzmzni ckuntinh, nemely, entroly is no lobger extensive. By am qqyensive entrkpy, I iewn, one where the following is true. The tntroly of a total system, is equal to the sum of jhe entropyes of 'each subsystem when considered in isolatiot', regeralews kd dome subsystems are "identical". This is equivalqht tp requiring thct the multiplicotj ps the total sistem, bz a product of the muptiplicyties of 'each fubsustem when considered in isilation'.
$$
\Omega_\next{rotal} = \prod_i \Omega_\cext{subsysteo $i$}
$$
Tne muktiplicity we have been usihg so far, cpearly dosr not satisfy thks iroparty, we had to multiply by a multinimiak coeffkciemt to det the cogrect total system multiolicijy;
$$
\Omegd_\text{total} \neq \Omega_\text{left box}\times\Omege_\vext{right box} \,,
$$
wverv $\Omega_\tert{left box}$ is the mtltiplicity of the leyt box, when the meft boe is considewed in isolathln, and similerly for $\Jmegq_\texr{right cux}$.
There is a nstural couuection bwtween how we have begn calculating muotiklucities so far, ana cjrgecv Bolesmann countitg. Bh rdsrraneing tht ewjatipn relating the totan syatem multiplicity (nenore the eivider ras removed) amd the multiplicitles oh the ueft snd right boxes (when considered ih isolatiln), re find
\begin{dathcr*}
\Omgga_\text{sys}
    = \Omega(V_1, N_1)\Omega(V_2, N_2) \times  \frac{(N_1+N_2)!}{N_1!N_2!} \\
\frec{\Omega_\text{sys}}{(N_
the divider. As a side note, we've that and $\Omega_\text{sys}'$ equal when ignoring should expect them to equal exactly either. removing the divider, each box contained definite number of particles, but after the divider is removed, the particle number each box will be variable. Thus, the Gibbs paradox is resolved without needing use Boltzmann If "naive" approach to calculating the multiplicity was correct all along, why should we still insist on by $N!$ to account for indistinguishability? How is by $N!$ even justified? is one downside to not correct counting, namely, is longer By an extensive I mean, one where the following is true. The entropy of a total system, is equal to sum of of 'each when in regardless if some "identical". This is equivalent to requiring of the total system, be a product of multiplicities of subsystem when considered in isolation'. $$ = \prod_i \Omega_\text{subsystem $i$} $$ The multiplicity we been using so far, clearly does not satisfy this property, we had to multiply by coefficient to get the total system multiplicity; \Omega_\text{total} \Omega_\text{left box} $$ where box}$ is the multiplicity of the left box, when the left is considered in isolation, and similarly for $\Omega_\text{right box}$. There natural between how we been calculating multiplicities so and Boltzmann counting. By rearranging relating total the was and the multiplicities of left and right boxes (when in isolation), we find N_2) \times \frac{(N_1+N_2)!}{N_1!N_2!} \\ \frac{\Omega_\text{sys}}{(N_
 the divider.
As a side note, we've Only shown tHat $\OmEga_\TexT{sYs}$ anD $\OmeGa_\text{sys}'$ are eqUAl whEn ignoring low order termS. ActuAlLY, we sHOuLd not Expect tHEm TO Be eQuAl ExaCtLY eIther. befOre remoVing the divIdeR, eAch box contaiNEd A definite nUmbEr of particleS, buT after ThE diVIder iS reMoved, The parTIcle nuMber in eacH bOX will bE VariablE.
tHuS, the gibbs paradox is resOLvED without needinG to use CoRReCT bolTzmAnn countinG. IF the "nAIve" apprOAcH TO CalCUlating the mulTiplicity waS CorRect alL aLonG, Why shoUld we StILl iNsist on diviDing By $N!$ to accoUnt for INdistinGUishabiLity? HoW is DivIsioN By $n!$ eVen JuSTifIEd?
theRE is One downsIdE tO not uSing CORREct BOltZmanN counTing, namely, entRopY is nO LonGer exTensiVe. By An ExtenSive enTropy, i mEan, one where the fOlloWing is truE. ThE eNtrOpY of a tOTal sysTem, Is eQual to tHe sum of THe eNtROPIeS of 'each subsystem whEn CONsIdered in IsolatIOn', ReGArdless iF sOme SubsYSTems aRe "idENtIcal". This Is equiVAlEnT to requIrIng thaT tHe mUltIplicITy of The totAl system, Be a prODuct of the multiPLicities of 'eacH SuBSYsTEm whEn cOnsidered in IsolATion'.
$$
\omegA_\TeXt{tOTal} = \prOd_i \OmEgA_\TeXT{subsystem $i$}
$$
The multiPlIcity wE have Been using so faR, clearly doES NOt satisfY thiS PrOPerty, we had to muLtiplY by a multinOMial coefFicieNt to get tHe correct TOTal systeM muLtiPliCitY;
$$
\oMeGa_\text{total} \neQ \oMega_\TeXt{left bOx}\tImes\OmeGa_\tExt{RigHt bOx} \,,
$$
Where $\OmegA_\text{lefT bOx}$ Is ThE muLtiplICity of thE lEft BoX, whEn the LEft box Is conSideReD iN IsoLation, aND sIMIlarLy FoR $\OmeGa_\tExT{righT box}$.
tHerE is a natUral conneCtiON betWeEn How we haVe been calculaTiNg multipliCiTieS so far, AND correct boltzmann counting. By rearRAnging tHe eQuatiOn reLating the TotAl systEm mULtipliCity (beFore tHe DivIDEr was REMoVed) AnD the multipLICitIes of ThE lefT and rigHt boxes (when consideREd iN isolation), we fInd
\BegiN{GAtHer*}
\oMeGA_\teXt{SYs}
    = \OMEGa(V_1, N_1)\Omega(V_2, N_2) \timeS  \frac{(N_1+N_2)!}{N_1!N_2!} \\
\fRaC{\omEga_\text{sys}}{(n_
 the divider.
As a side no te, we'veonlysho wnth at $ \Ome ga_\text{sys}$ and$\Omega_\text{sys}'$ a re eq ua l  whe n  i gnori ng lowo rd e r  te rm s.  Ac tu a ll y, we  sh ould no t expect t hem  t o be equal e x ac tly either . B efore removi ngthe di vi der ,  each  bo x con tained a defi nite numb er of par t icles,b u tafte r the divider isr em o ved, the parti cle nu mb e ri n  ea chbox will b evaria b le.
Thu s ,t h e  Gi b bs paradox is  resolved w i tho ut nee di ngt o usecorre ct Bol tzmann coun ting . If the"naive "  approa c h to ca lculat ing  th e mu l ti pl ici ty was co rre c t a ll along ,wh y sho uldw e s till  in sist  on d ividing by $N !$to a c cou nt fo r ind isti ng uisha bility ? How  i s division by $ N!$even just ifi ed ?
T he re is one do wns ide  to not  usingc orr ec t B ol tzmann counting, n am e l y,  entropy  is no lo ng e r extens iv e.By a n exten sive en tropy, I  mean, on ewhere t he  follo wi ngistrue. Theentrop y of a t otals ystem, is equa l  to the sum o f  t h e  e n trop ies  of 'each s ubsy s temwhen co nsi d eredin is ol a ti o n', regardless if s om e subs ystem s are "identi cal". This i s  equival entt or equiring thatthe m ultiplicit y  of thetotal  system,  be a pro d u ct of th e m ult ipl ici t i es  of 'each sub s y stem  w hen con sid ered in  is ola tio n'. 
$ $
\Omega_ \text{to ta l}  =  \ pro d_i \ O mega_\te xt {su bs yst em $i $ }
$$
T he mu ltip li ci t y w e haveb ee n usin gso  far , c le arlydoes not  satisf y this pr ope r ty,we  h ad to m ultiply by amu ltinomialco eff icient t o get th e correct total systemm ultipli cit y;
$$ 
\Om ega_\text {to tal} \ neq \Omega _\text {left  b ox} \ t imes\ O m eg a_\ te xt{right b o x } \ ,,
$$ 
w here  $\Omeg a_\text{left box}$ isthe multiplic ity  oft h elef t  b o x,wh e n t h e  left box is co nsidered i ni so lation, an d  si mi larly f or $\Om ega_\ t ext{rig ht box}$. 
There is  a  nat u r alconnection  between  how we h a ve be e ncalcu lat ing mu lt ipl iciti es sof ar,  andcorrec tBoltzm ann c ou nting. B y rearranging the equat ion re latin g t he totalsys t emmultiplic ity(before th e d ivi der w asr emove d) a n dthe multi plic i ties of t h elef t an d right box e s (wh en co nsi d ered i n is olation), we find \begin{gather* }
\O m e ga_ \te x t{sy s} 
    = \Omega( V_1 ,N _ 1)\Omega (V _2, N_2) \t imes  \f ra c {(N_1 +N_2)! }{N_1! N_2!} \ \ \f r ac{\Om ega_ \te xt{sys}}{ (N_ 
 the_divider.
As a_side note, we've only_shown that_$\Omega_\text{sys}$_and $\Omega_\text{sys}'$_are_equal when ignoring_low order terms._Actually, we should not_expect them to_be_equal exactly either. Before removing the divider, each box contained a definite number of_particles,_but after_the_divider_is removed, the particle number_in each box will be_variable.
Thus, the_Gibbs paradox is resolved without needing to use_correct_Boltzmann counting. If_the "naive" approach to calculating the multiplicity was correct_all along, why should we still_insist on dividing_by_$N!$_to account for indistinguishability?_How is division by $N!$ even_justified?
There is one downside to not_using correct Boltzmann counting, namely, entropy is_no longer extensive. By an extensive_entropy, I mean, one where_the following_is true. The entropy of_a total system,_is equal_to the sum_of the entropies of 'each subsystem_when considered in_isolation', regardless if some subsystems are_"identical"._This is equivalent_to_requiring_that the_multiplicity of the_total_system, be_a_product of the multiplicities of 'each_subsystem_when considered in isolation'.
$$
\Omega_\text{total} = \prod_i \Omega_\text{subsystem_$i$}
$$
The multiplicity we have_been_using so far, clearly_does not satisfy this property,_we had to multiply by a_multinomial coefficient_to get_the correct total system multiplicity;
$$
\Omega_\text{total} \neq \Omega_\text{left box}\times\Omega_\text{right box} \,,
$$
where $\Omega_\text{left_box}$ is the multiplicity of the_left box, when the_left box_is_considered in isolation,_and_similarly for_$\Omega_\text{right box}$.
There is a natural connection between_how we_have been calculating multiplicities so far,_and correct Boltzmann counting._By_rearranging the equation relating the total_system multiplicity (before the divider was_removed) and the multiplicities of_the_left_and right boxes (when considered_in isolation), we find
\begin{gather*}
\Omega_\text{sys}
  _ = \Omega(V_1,_N_1)\Omega(V_2, N_2) \times  \frac{(N_1+N_2)!}{N_1!N_2!} \\
\frac{\Omega_\text{sys}}{(N_