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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File:DeusFollowingSea.jpg

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File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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