In how many ways can a committee be formed if a certain male and a certain female cannot serve together?...
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A class consists of $14$ males and $12$ females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of $6$
men and $4$ women be chosen from the class?
My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?
combinatorics combinations
asked Nov 12 at 10:59
Georges Shimmon
41
closed as off-topic by 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk Nov 13 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-2
down vote
favorite
A class consists of $14$ males and $12$ females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of $6$
men and $4$ women be chosen from the class?
My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?
combinatorics combinations
asked Nov 12 at 10:59
Georges Shimmon
41
closed as off-topic by 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk Nov 13 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk
If this question can be reworded to fit the rules in the help center, please edit the question.
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 12 at 11:01
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 12 at 11:01
ok no worries ill add my working now, thanks for the feedback
– Georges Shimmon
Nov 12 at 11:02
Are you leaving something out of the problem statement? It seems we need to know the numbers of men and women in the class.
– awkward
Nov 12 at 15:30
1
What you have counted is the number of committees with neither male A nor female B. What you have not counted is the number of committees on which male A serves but female B does not and the number of committees on which female B serves but male A does not.
– N. F. Taussig
Nov 12 at 20:18
|
show 1 more comment
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
A class consists of $14$ males and $12$ females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of $6$
men and $4$ women be chosen from the class?
My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?
combinatorics combinations
asked Nov 12 at 10:59
Georges Shimmon
41
A class consists of $14$ males and $12$ females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of $6$
men and $4$ women be chosen from the class?
My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?
combinatorics combinations
combinatorics combinations
asked Nov 12 at 10:59
Georges Shimmon
41
asked Nov 12 at 10:59
Georges Shimmon
41
edited Nov 16 at 0:34
asked Nov 12 at 10:59
Georges Shimmon
41
asked Nov 12 at 10:59
Georges Shimmon
41
asked Nov 12 at 10:59
Georges Shimmon
41
41
closed as off-topic by 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk Nov 13 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk Nov 13 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, max_zorn, Rushabh Mehta, Shailesh, ArsenBerk
If this question can be reworded to fit the rules in the help center, please edit the question.
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 12 at 11:01
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 12 at 11:01
ok no worries ill add my working now, thanks for the feedback
– Georges Shimmon
Nov 12 at 11:02
Are you leaving something out of the problem statement? It seems we need to know the numbers of men and women in the class.
– awkward
Nov 12 at 15:30
1
What you have counted is the number of committees with neither male A nor female B. What you have not counted is the number of committees on which male A serves but female B does not and the number of committees on which female B serves but male A does not.
– N. F. Taussig
Nov 12 at 20:18
|
show 1 more comment
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 12 at 11:01
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 12 at 11:01
ok no worries ill add my working now, thanks for the feedback
– Georges Shimmon
Nov 12 at 11:02
Are you leaving something out of the problem statement? It seems we need to know the numbers of men and women in the class.
– awkward
Nov 12 at 15:30
1
What you have counted is the number of committees with neither male A nor female B. What you have not counted is the number of committees on which male A serves but female B does not and the number of committees on which female B serves but male A does not.
– N. F. Taussig
Nov 12 at 20:18
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 12 at 11:01
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 12 at 11:01
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 12 at 11:01
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 12 at 11:01
ok no worries ill add my working now, thanks for the feedback
– Georges Shimmon
Nov 12 at 11:02
ok no worries ill add my working now, thanks for the feedback
– Georges Shimmon
Nov 12 at 11:02
Are you leaving something out of the problem statement? It seems we need to know the numbers of men and women in the class.
– awkward
Nov 12 at 15:30
Are you leaving something out of the problem statement? It seems we need to know the numbers of men and women in the class.
– awkward
Nov 12 at 15:30
1
1
What you have counted is the number of committees with neither male A nor female B. What you have not counted is the number of committees on which male A serves but female B does not and the number of committees on which female B serves but male A does not.
– N. F. Taussig
Nov 12 at 20:18
What you have counted is the number of committees with neither male A nor female B. What you have not counted is the number of committees on which male A serves but female B does not and the number of committees on which female B serves but male A does not.
– N. F. Taussig
Nov 12 at 20:18
|
show 1 more comment
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Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 12 at 11:01
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 12 at 11:01
ok no worries ill add my working now, thanks for the feedback
– Georges Shimmon
Nov 12 at 11:02
Are you leaving something out of the problem statement? It seems we need to know the numbers of men and women in the class.
– awkward
Nov 12 at 15:30
1
What you have counted is the number of committees with neither male A nor female B. What you have not counted is the number of committees on which male A serves but female B does not and the number of committees on which female B serves but male A does not.
– N. F. Taussig
Nov 12 at 20:18