Combinatorics- Why is this approach wrong?












2












$begingroup$


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
    Jan 22 at 12:23










  • $begingroup$
    @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
    Jan 22 at 12:25












  • $begingroup$
    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
    Jan 22 at 12:28










  • $begingroup$
    @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
    Jan 22 at 12:29










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    Jan 22 at 12:32
















2












$begingroup$


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
    Jan 22 at 12:23










  • $begingroup$
    @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
    Jan 22 at 12:25












  • $begingroup$
    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
    Jan 22 at 12:28










  • $begingroup$
    @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
    Jan 22 at 12:29










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    Jan 22 at 12:32














2












2








2


0



$begingroup$


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 Jan 22 at 12:31









Naman Kumar

23413




23413










asked Jan 22 at 12:18









Sunil Kumar JhaSunil Kumar Jha

1387




1387












  • $begingroup$
    What was the reasoning for your " first try" answer?
    $endgroup$
    – coffeemath
    Jan 22 at 12:23










  • $begingroup$
    @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
    Jan 22 at 12:25












  • $begingroup$
    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
    Jan 22 at 12:28










  • $begingroup$
    @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
    Jan 22 at 12:29










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    Jan 22 at 12:32


















  • $begingroup$
    What was the reasoning for your " first try" answer?
    $endgroup$
    – coffeemath
    Jan 22 at 12:23










  • $begingroup$
    @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
    Jan 22 at 12:25












  • $begingroup$
    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
    Jan 22 at 12:28










  • $begingroup$
    @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
    Jan 22 at 12:29










  • $begingroup$
    I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
    $endgroup$
    – coffeemath
    Jan 22 at 12:32
















$begingroup$
What was the reasoning for your " first try" answer?
$endgroup$
– coffeemath
Jan 22 at 12:23




$begingroup$
What was the reasoning for your " first try" answer?
$endgroup$
– coffeemath
Jan 22 at 12:23












$begingroup$
@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
Jan 22 at 12:25






$begingroup$
@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
Jan 22 at 12:25














$begingroup$
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
Jan 22 at 12:28




$begingroup$
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
Jan 22 at 12:28












$begingroup$
@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
Jan 22 at 12:29




$begingroup$
@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
Jan 22 at 12:29












$begingroup$
I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
$endgroup$
– coffeemath
Jan 22 at 12:32




$begingroup$
I think the problem is that when you choose 2M,3W you have to say specifically which M and W were chosen.
$endgroup$
– coffeemath
Jan 22 at 12:32










2 Answers
2






active

oldest

votes


















4












$begingroup$

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
    Jan 22 at 12:34












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    Jan 22 at 12:35



















2












$begingroup$

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
    Jan 22 at 12:37











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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
    Jan 22 at 12:34












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    Jan 22 at 12:35
















4












$begingroup$

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
    Jan 22 at 12:34












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    Jan 22 at 12:35














4












4








4





$begingroup$

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 Jan 22 at 12:35

























answered Jan 22 at 12:28









Robert ZRobert Z

98.5k1068139




98.5k1068139












  • $begingroup$
    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
    Jan 22 at 12:34












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    Jan 22 at 12:35


















  • $begingroup$
    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
    Jan 22 at 12:34












  • $begingroup$
    @SunilKumarJha Yes, exactly.
    $endgroup$
    – Robert Z
    Jan 22 at 12:35
















$begingroup$
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
Jan 22 at 12:34






$begingroup$
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
Jan 22 at 12:34














$begingroup$
@SunilKumarJha Yes, exactly.
$endgroup$
– Robert Z
Jan 22 at 12:35




$begingroup$
@SunilKumarJha Yes, exactly.
$endgroup$
– Robert Z
Jan 22 at 12:35











2












$begingroup$

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
    Jan 22 at 12:37
















2












$begingroup$

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
    Jan 22 at 12:37














2












2








2





$begingroup$

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 Jan 22 at 12:32









drhabdrhab

102k545136




102k545136












  • $begingroup$
    Thanks, i understood .
    $endgroup$
    – Sunil Kumar Jha
    Jan 22 at 12:37


















  • $begingroup$
    Thanks, i understood .
    $endgroup$
    – Sunil Kumar Jha
    Jan 22 at 12:37
















$begingroup$
Thanks, i understood .
$endgroup$
– Sunil Kumar Jha
Jan 22 at 12:37




$begingroup$
Thanks, i understood .
$endgroup$
– Sunil Kumar Jha
Jan 22 at 12:37


















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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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