Combinatorics- Why is this approach wrong?












1












$begingroup$


In how many ways a committee of 6 members to be formed from 8 men and 5 women such that there are at least 2 men and 3 women?



Let men be $m$ and women be $f$.



I got the answer in second try using cases ($4f+2m$ or $3f+3m$) but in my first try I did something like this: ${8 choose 2}{5 choose 3}{8+5-5 choose 1}$.



According to me this should have given the right answer but it didn't.



Can someone please indicate what is the mistake I have made?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    What was the reasoning for your " first try" answer?
    $endgroup$
    – coffeemath
    58 mins ago










  • $begingroup$
    @coffeemath As the committee should have at least 2 men and 3 women i could do that in C(8,2)*C(5,3) and for last one person which can be either male or female, i have got 8 person to select from.So,C(8,1)
    $endgroup$
    – Sunil Kumar Jha
    55 mins ago












  • $begingroup$
    That way starts with exactly 2 men and 3 women. But that's only 5 on committee. If one goes on from there and puts another person on committee, it goes against first choice.
    $endgroup$
    – coffeemath
    53 mins ago










  • $begingroup$
    @coffeemath "If one goes on from there and puts another person on committee, it goes against first choice" can you explain this more?
    $endgroup$
    – Sunil Kumar Jha
    51 mins ago










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    49 mins ago
















1












$begingroup$


In how many ways a committee of 6 members to be formed from 8 men and 5 women such that there are at least 2 men and 3 women?



Let men be $m$ and women be $f$.



I got the answer in second try using cases ($4f+2m$ or $3f+3m$) but in my first try I did something like this: ${8 choose 2}{5 choose 3}{8+5-5 choose 1}$.



According to me this should have given the right answer but it didn't.



Can someone please indicate what is the mistake I have made?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    What was the reasoning for your " first try" answer?
    $endgroup$
    – coffeemath
    58 mins ago










  • $begingroup$
    @coffeemath As the committee should have at least 2 men and 3 women i could do that in C(8,2)*C(5,3) and for last one person which can be either male or female, i have got 8 person to select from.So,C(8,1)
    $endgroup$
    – Sunil Kumar Jha
    55 mins ago












  • $begingroup$
    That way starts with exactly 2 men and 3 women. But that's only 5 on committee. If one goes on from there and puts another person on committee, it goes against first choice.
    $endgroup$
    – coffeemath
    53 mins ago










  • $begingroup$
    @coffeemath "If one goes on from there and puts another person on committee, it goes against first choice" can you explain this more?
    $endgroup$
    – Sunil Kumar Jha
    51 mins ago










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    49 mins ago














1












1








1


1



$begingroup$


In how many ways a committee of 6 members to be formed from 8 men and 5 women such that there are at least 2 men and 3 women?



Let men be $m$ and women be $f$.



I got the answer in second try using cases ($4f+2m$ or $3f+3m$) but in my first try I did something like this: ${8 choose 2}{5 choose 3}{8+5-5 choose 1}$.



According to me this should have given the right answer but it didn't.



Can someone please indicate what is the mistake I have made?






combinatorics







share|cite|improve this question











$endgroup$




In how many ways a committee of 6 members to be formed from 8 men and 5 women such that there are at least 2 men and 3 women?



Let men be $m$ and women be $f$.



I got the answer in second try using cases ($4f+2m$ or $3f+3m$) but in my first try I did something like this: ${8 choose 2}{5 choose 3}{8+5-5 choose 1}$.



According to me this should have given the right answer but it didn't.



Can someone please indicate what is the mistake I have made?






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 49 mins ago









Naman Kumar

5810




5810










asked 1 hour ago









Sunil Kumar JhaSunil Kumar Jha

1336




1336












  • $begingroup$
    What was the reasoning for your " first try" answer?
    $endgroup$
    – coffeemath
    58 mins ago










  • $begingroup$
    @coffeemath As the committee should have at least 2 men and 3 women i could do that in C(8,2)*C(5,3) and for last one person which can be either male or female, i have got 8 person to select from.So,C(8,1)
    $endgroup$
    – Sunil Kumar Jha
    55 mins ago












  • $begingroup$
    That way starts with exactly 2 men and 3 women. But that's only 5 on committee. If one goes on from there and puts another person on committee, it goes against first choice.
    $endgroup$
    – coffeemath
    53 mins ago










  • $begingroup$
    @coffeemath "If one goes on from there and puts another person on committee, it goes against first choice" can you explain this more?
    $endgroup$
    – Sunil Kumar Jha
    51 mins ago










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    49 mins ago


















  • $begingroup$
    What was the reasoning for your " first try" answer?
    $endgroup$
    – coffeemath
    58 mins ago










  • $begingroup$
    @coffeemath As the committee should have at least 2 men and 3 women i could do that in C(8,2)*C(5,3) and for last one person which can be either male or female, i have got 8 person to select from.So,C(8,1)
    $endgroup$
    – Sunil Kumar Jha
    55 mins ago












  • $begingroup$
    That way starts with exactly 2 men and 3 women. But that's only 5 on committee. If one goes on from there and puts another person on committee, it goes against first choice.
    $endgroup$
    – coffeemath
    53 mins ago










  • $begingroup$
    @coffeemath "If one goes on from there and puts another person on committee, it goes against first choice" can you explain this more?
    $endgroup$
    – Sunil Kumar Jha
    51 mins ago










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    49 mins ago
















$begingroup$
What was the reasoning for your " first try" answer?
$endgroup$
– coffeemath
58 mins ago




$begingroup$
What was the reasoning for your " first try" answer?
$endgroup$
– coffeemath
58 mins ago












$begingroup$
@coffeemath As the committee should have at least 2 men and 3 women i could do that in C(8,2)*C(5,3) and for last one person which can be either male or female, i have got 8 person to select from.So,C(8,1)
$endgroup$
– Sunil Kumar Jha
55 mins ago






$begingroup$
@coffeemath As the committee should have at least 2 men and 3 women i could do that in C(8,2)*C(5,3) and for last one person which can be either male or female, i have got 8 person to select from.So,C(8,1)
$endgroup$
– Sunil Kumar Jha
55 mins ago














$begingroup$
That way starts with exactly 2 men and 3 women. But that's only 5 on committee. If one goes on from there and puts another person on committee, it goes against first choice.
$endgroup$
– coffeemath
53 mins ago




$begingroup$
That way starts with exactly 2 men and 3 women. But that's only 5 on committee. If one goes on from there and puts another person on committee, it goes against first choice.
$endgroup$
– coffeemath
53 mins ago












$begingroup$
@coffeemath "If one goes on from there and puts another person on committee, it goes against first choice" can you explain this more?
$endgroup$
– Sunil Kumar Jha
51 mins ago




$begingroup$
@coffeemath "If one goes on from there and puts another person on committee, it goes against first choice" can you explain this more?
$endgroup$
– Sunil Kumar Jha
51 mins ago












$begingroup$
I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
$endgroup$
– coffeemath
49 mins ago




$begingroup$
I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
$endgroup$
– coffeemath
49 mins ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

By using the formula
$$binom{8}{2}cdot binom{5}{3}cdot binom{8+5-5}{1}$$
you are count the same committee more than one time.
Let $m_1,dots, m_8$ the male members and $f_1,dots, f_5$ the female members.
Then the committee $m_1m_2f_1f_2f_3f_4$ is counted one time when you first choose $f_1f_2f_3$ and then $f_4$ and a second time when you first choose $f_2f_3f_4$ and then $f_1$.



On the other hand, as you already noted, if we consider the two possible cases $2m+4f$ or $3m+3f$, we obtain the formula
$$binom{8}{2}cdot binom{5}{4}+binom{8}{3}cdot binom{5}{3}$$
which counts every committee one and only one time.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times?
    $endgroup$
    – Sunil Kumar Jha
    46 mins ago












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    45 mins ago



















1












$begingroup$

I suspect factor $C(8,2)$ stands for the number of ways $2$ men can be selected out of $8$ and similarly $C(5,3)$ for the number of ways $3$ women can be selected out of $5$.



Then you probabibly reason: $8+5-5=8$ persons are left and out of them $1$ must be chosen, leading to factor $C(8,1)$.



This however gives multiple counting.



Suppose Albert and Bob are chosen at first hand and Carl is chosen as the one taken out the remaining $8$.



This leads to the same mail committee members in the case where Albert and Carl are chosen at first hand and Bob is chosen as the one taken out of the remaining $8$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, i understood .
    $endgroup$
    – Sunil Kumar Jha
    44 mins ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083088%2fcombinatorics-why-is-this-approach-wrong%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

By using the formula
$$binom{8}{2}cdot binom{5}{3}cdot binom{8+5-5}{1}$$
you are count the same committee more than one time.
Let $m_1,dots, m_8$ the male members and $f_1,dots, f_5$ the female members.
Then the committee $m_1m_2f_1f_2f_3f_4$ is counted one time when you first choose $f_1f_2f_3$ and then $f_4$ and a second time when you first choose $f_2f_3f_4$ and then $f_1$.



On the other hand, as you already noted, if we consider the two possible cases $2m+4f$ or $3m+3f$, we obtain the formula
$$binom{8}{2}cdot binom{5}{4}+binom{8}{3}cdot binom{5}{3}$$
which counts every committee one and only one time.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times?
    $endgroup$
    – Sunil Kumar Jha
    46 mins ago












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    45 mins ago
















3












$begingroup$

By using the formula
$$binom{8}{2}cdot binom{5}{3}cdot binom{8+5-5}{1}$$
you are count the same committee more than one time.
Let $m_1,dots, m_8$ the male members and $f_1,dots, f_5$ the female members.
Then the committee $m_1m_2f_1f_2f_3f_4$ is counted one time when you first choose $f_1f_2f_3$ and then $f_4$ and a second time when you first choose $f_2f_3f_4$ and then $f_1$.



On the other hand, as you already noted, if we consider the two possible cases $2m+4f$ or $3m+3f$, we obtain the formula
$$binom{8}{2}cdot binom{5}{4}+binom{8}{3}cdot binom{5}{3}$$
which counts every committee one and only one time.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times?
    $endgroup$
    – Sunil Kumar Jha
    46 mins ago












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    45 mins ago














3












3








3





$begingroup$

By using the formula
$$binom{8}{2}cdot binom{5}{3}cdot binom{8+5-5}{1}$$
you are count the same committee more than one time.
Let $m_1,dots, m_8$ the male members and $f_1,dots, f_5$ the female members.
Then the committee $m_1m_2f_1f_2f_3f_4$ is counted one time when you first choose $f_1f_2f_3$ and then $f_4$ and a second time when you first choose $f_2f_3f_4$ and then $f_1$.



On the other hand, as you already noted, if we consider the two possible cases $2m+4f$ or $3m+3f$, we obtain the formula
$$binom{8}{2}cdot binom{5}{4}+binom{8}{3}cdot binom{5}{3}$$
which counts every committee one and only one time.






share|cite|improve this answer











$endgroup$



By using the formula
$$binom{8}{2}cdot binom{5}{3}cdot binom{8+5-5}{1}$$
you are count the same committee more than one time.
Let $m_1,dots, m_8$ the male members and $f_1,dots, f_5$ the female members.
Then the committee $m_1m_2f_1f_2f_3f_4$ is counted one time when you first choose $f_1f_2f_3$ and then $f_4$ and a second time when you first choose $f_2f_3f_4$ and then $f_1$.



On the other hand, as you already noted, if we consider the two possible cases $2m+4f$ or $3m+3f$, we obtain the formula
$$binom{8}{2}cdot binom{5}{4}+binom{8}{3}cdot binom{5}{3}$$
which counts every committee one and only one time.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 46 mins ago

























answered 52 mins ago









Robert ZRobert Z

95.2k1063134




95.2k1063134












  • $begingroup$
    okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times?
    $endgroup$
    – Sunil Kumar Jha
    46 mins ago












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    45 mins ago


















  • $begingroup$
    okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times?
    $endgroup$
    – Sunil Kumar Jha
    46 mins ago












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    45 mins ago
















$begingroup$
okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times?
$endgroup$
– Sunil Kumar Jha
46 mins ago






$begingroup$
okay i understood what you are saying.Like if there are 8 men and 5 women and if i have selected 2 men then i am left with 6 men and when i am selecting last person, i am forming same committee again in few cases.Is this counts every possible committee equal number of times or different number of times?
$endgroup$
– Sunil Kumar Jha
46 mins ago














$begingroup$
@SunilKumarJha Yes, exactly.
$endgroup$
– Robert Z
45 mins ago




$begingroup$
@SunilKumarJha Yes, exactly.
$endgroup$
– Robert Z
45 mins ago











1












$begingroup$

I suspect factor $C(8,2)$ stands for the number of ways $2$ men can be selected out of $8$ and similarly $C(5,3)$ for the number of ways $3$ women can be selected out of $5$.



Then you probabibly reason: $8+5-5=8$ persons are left and out of them $1$ must be chosen, leading to factor $C(8,1)$.



This however gives multiple counting.



Suppose Albert and Bob are chosen at first hand and Carl is chosen as the one taken out the remaining $8$.



This leads to the same mail committee members in the case where Albert and Carl are chosen at first hand and Bob is chosen as the one taken out of the remaining $8$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, i understood .
    $endgroup$
    – Sunil Kumar Jha
    44 mins ago
















1












$begingroup$

I suspect factor $C(8,2)$ stands for the number of ways $2$ men can be selected out of $8$ and similarly $C(5,3)$ for the number of ways $3$ women can be selected out of $5$.



Then you probabibly reason: $8+5-5=8$ persons are left and out of them $1$ must be chosen, leading to factor $C(8,1)$.



This however gives multiple counting.



Suppose Albert and Bob are chosen at first hand and Carl is chosen as the one taken out the remaining $8$.



This leads to the same mail committee members in the case where Albert and Carl are chosen at first hand and Bob is chosen as the one taken out of the remaining $8$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, i understood .
    $endgroup$
    – Sunil Kumar Jha
    44 mins ago














1












1








1





$begingroup$

I suspect factor $C(8,2)$ stands for the number of ways $2$ men can be selected out of $8$ and similarly $C(5,3)$ for the number of ways $3$ women can be selected out of $5$.



Then you probabibly reason: $8+5-5=8$ persons are left and out of them $1$ must be chosen, leading to factor $C(8,1)$.



This however gives multiple counting.



Suppose Albert and Bob are chosen at first hand and Carl is chosen as the one taken out the remaining $8$.



This leads to the same mail committee members in the case where Albert and Carl are chosen at first hand and Bob is chosen as the one taken out of the remaining $8$.






share|cite|improve this answer









$endgroup$



I suspect factor $C(8,2)$ stands for the number of ways $2$ men can be selected out of $8$ and similarly $C(5,3)$ for the number of ways $3$ women can be selected out of $5$.



Then you probabibly reason: $8+5-5=8$ persons are left and out of them $1$ must be chosen, leading to factor $C(8,1)$.



This however gives multiple counting.



Suppose Albert and Bob are chosen at first hand and Carl is chosen as the one taken out the remaining $8$.



This leads to the same mail committee members in the case where Albert and Carl are chosen at first hand and Bob is chosen as the one taken out of the remaining $8$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 49 mins ago









drhabdrhab

99.4k544130




99.4k544130












  • $begingroup$
    Thanks, i understood .
    $endgroup$
    – Sunil Kumar Jha
    44 mins ago


















  • $begingroup$
    Thanks, i understood .
    $endgroup$
    – Sunil Kumar Jha
    44 mins ago
















$begingroup$
Thanks, i understood .
$endgroup$
– Sunil Kumar Jha
44 mins ago




$begingroup$
Thanks, i understood .
$endgroup$
– Sunil Kumar Jha
44 mins ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3083088%2fcombinatorics-why-is-this-approach-wrong%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ponta tanko

Image
Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
Read more

Tantalo (mitologio)

Image
Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
Read more

Erzsébet Schaár

Image
Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
Read more