a committee of 3 women and 4 men is to be formed from 6 women and 7 men [duplicate]
This question already has an answer here:
Selecting a security team that includes at most one of the oldest man or oldest woman
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a committee of 3 women and 4 men is to be formed from 6 women and 7 men. How many ways committee can be formed if it can only include at most 1 of the youngest woman or youngest man.
case 1: neither are included -> 5C3 X 6C2
case 2: only youngest woman included -> 5C3 X 7C4
case 3: only youngest man included -> 6C3 X 6C3
i then add all of those up together. is this right?
combinatorics
asked Nov 27 at 11:28
Erikien
494
marked as duplicate by lulu, N. F. Taussig combinatorics
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Nov 27 at 12:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Selecting a security team that includes at most one of the oldest man or oldest woman
1 answer
a committee of 3 women and 4 men is to be formed from 6 women and 7 men. How many ways committee can be formed if it can only include at most 1 of the youngest woman or youngest man.
case 1: neither are included -> 5C3 X 6C2
case 2: only youngest woman included -> 5C3 X 7C4
case 3: only youngest man included -> 6C3 X 6C3
i then add all of those up together. is this right?
combinatorics
asked Nov 27 at 11:28
Erikien
494
marked as duplicate by lulu, N. F. Taussig combinatorics
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Nov 27 at 12:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman.
– lulu
Nov 27 at 11:37
This is essentially a duplicate of your prior question....study the answers you were given to that one and apply them to this (minor) variant.
– lulu
Nov 27 at 12:01
add a comment |
This question already has an answer here:
Selecting a security team that includes at most one of the oldest man or oldest woman
1 answer
a committee of 3 women and 4 men is to be formed from 6 women and 7 men. How many ways committee can be formed if it can only include at most 1 of the youngest woman or youngest man.
case 1: neither are included -> 5C3 X 6C2
case 2: only youngest woman included -> 5C3 X 7C4
case 3: only youngest man included -> 6C3 X 6C3
i then add all of those up together. is this right?
combinatorics
asked Nov 27 at 11:28
Erikien
494
This question already has an answer here:
Selecting a security team that includes at most one of the oldest man or oldest woman
1 answer
a committee of 3 women and 4 men is to be formed from 6 women and 7 men. How many ways committee can be formed if it can only include at most 1 of the youngest woman or youngest man.
case 1: neither are included -> 5C3 X 6C2
case 2: only youngest woman included -> 5C3 X 7C4
case 3: only youngest man included -> 6C3 X 6C3
i then add all of those up together. is this right?
This question already has an answer here:
Selecting a security team that includes at most one of the oldest man or oldest woman
1 answer
combinatorics
combinatorics
asked Nov 27 at 11:28
Erikien
494
asked Nov 27 at 11:28
Erikien
494
asked Nov 27 at 11:28
Erikien
494
asked Nov 27 at 11:28
Erikien
494
asked Nov 27 at 11:28
Erikien
494
494
marked as duplicate by lulu, N. F. Taussig combinatorics
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Nov 27 at 12:17
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Nov 27 at 12:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman.
– lulu
Nov 27 at 11:37
This is essentially a duplicate of your prior question....study the answers you were given to that one and apply them to this (minor) variant.
– lulu
Nov 27 at 12:01
add a comment |
All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman.
– lulu
Nov 27 at 11:37
This is essentially a duplicate of your prior question....study the answers you were given to that one and apply them to this (minor) variant.
– lulu
Nov 27 at 12:01
All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman.
– lulu
Nov 27 at 11:37
All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman.
– lulu
Nov 27 at 11:37
This is essentially a duplicate of your prior question....study the answers you were given to that one and apply them to this (minor) variant.
– lulu
Nov 27 at 12:01
This is essentially a duplicate of your prior question....study the answers you were given to that one and apply them to this (minor) variant.
– lulu
Nov 27 at 12:01
add a comment |
2 Answers
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Without the age restriction, there would be $binom{6}{3}binom{7}{4}$. Of these, exactly $binom{5}{2}binom{6}{3}$ are illegal due to featuring the youngest member of each sex. This makes the number of legal solutions $binom{6}{3}(binom{7}{4}-binom{5}{2})=20times (35-10)=500$.
answered Nov 27 at 11:38
J.G.
22.5k22035
add a comment |
This doesn't seem to add up, in case 1 you're only choosing 5 committee members instead of 7.
In case 2 you're choosing 7 but should be choosing 6 since the youngest woman is included.
Case 3 looks correct to me.
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
add a comment |
2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Without the age restriction, there would be $binom{6}{3}binom{7}{4}$. Of these, exactly $binom{5}{2}binom{6}{3}$ are illegal due to featuring the youngest member of each sex. This makes the number of legal solutions $binom{6}{3}(binom{7}{4}-binom{5}{2})=20times (35-10)=500$.
answered Nov 27 at 11:38
J.G.
22.5k22035
add a comment |
Without the age restriction, there would be $binom{6}{3}binom{7}{4}$. Of these, exactly $binom{5}{2}binom{6}{3}$ are illegal due to featuring the youngest member of each sex. This makes the number of legal solutions $binom{6}{3}(binom{7}{4}-binom{5}{2})=20times (35-10)=500$.
answered Nov 27 at 11:38
J.G.
22.5k22035
add a comment |
Without the age restriction, there would be $binom{6}{3}binom{7}{4}$. Of these, exactly $binom{5}{2}binom{6}{3}$ are illegal due to featuring the youngest member of each sex. This makes the number of legal solutions $binom{6}{3}(binom{7}{4}-binom{5}{2})=20times (35-10)=500$.
answered Nov 27 at 11:38
J.G.
22.5k22035
Without the age restriction, there would be $binom{6}{3}binom{7}{4}$. Of these, exactly $binom{5}{2}binom{6}{3}$ are illegal due to featuring the youngest member of each sex. This makes the number of legal solutions $binom{6}{3}(binom{7}{4}-binom{5}{2})=20times (35-10)=500$.
answered Nov 27 at 11:38
J.G.
22.5k22035
answered Nov 27 at 11:38
J.G.
22.5k22035
answered Nov 27 at 11:38
J.G.
22.5k22035
answered Nov 27 at 11:38
J.G.
22.5k22035
22.5k22035
add a comment |
add a comment |
This doesn't seem to add up, in case 1 you're only choosing 5 committee members instead of 7.
In case 2 you're choosing 7 but should be choosing 6 since the youngest woman is included.
Case 3 looks correct to me.
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
add a comment |
This doesn't seem to add up, in case 1 you're only choosing 5 committee members instead of 7.
In case 2 you're choosing 7 but should be choosing 6 since the youngest woman is included.
Case 3 looks correct to me.
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
add a comment |
This doesn't seem to add up, in case 1 you're only choosing 5 committee members instead of 7.
In case 2 you're choosing 7 but should be choosing 6 since the youngest woman is included.
Case 3 looks correct to me.
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
This doesn't seem to add up, in case 1 you're only choosing 5 committee members instead of 7.
In case 2 you're choosing 7 but should be choosing 6 since the youngest woman is included.
Case 3 looks correct to me.
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
answered Nov 27 at 11:32
Olivier Moschetta
2,7761411
2,7761411
add a comment |
add a comment |
All three are wrong. For $1$ you are choosing five people, for $2$ some of your combinations include the youngest man, for $3$ some of your combinations include the youngest woman.
– lulu
Nov 27 at 11:37
This is essentially a duplicate of your prior question....study the answers you were given to that one and apply them to this (minor) variant.
– lulu
Nov 27 at 12:01