Mixing
Find the number of 5-member committiees which include at least two Republicans
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up vote
3
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favorite
There are 10 Republicans, 8 Democrats and 2 Independent legislators eligible for committiee membership.
How many 5-member committiees exist which include at least two Republicans?
My Work :
$C(20,5)-(C(10,5)+C(10,4))$
$(;All; 5-member; committiees;);-;(;committiees; with ;no; Republicans;+;committiees; with; one ;Republican ;)$
Explanation: There are $C(20,5)$ 5-member committiees which will be subtracted from sum of two numbers in order to achieve the answer; C(10,5) is number of 5-member committiees which include no Republicans (I've only counted Democrats and Independents ($8+2$)then found number of 5-member committiees ) and C(10,4) is number of 5-member committiees with one Republican (Like before I've excluded Republicans but counted 4-member committiees then added one Republican to each subset)
Book's Answer:
It's just said $C(10,2)C(18,3)$
combinatorics binomial-theorem
asked 1 hour ago
Abbas
897
add a comment |Â
up vote
3
down vote
favorite
There are 10 Republicans, 8 Democrats and 2 Independent legislators eligible for committiee membership.
How many 5-member committiees exist which include at least two Republicans?
My Work :
$C(20,5)-(C(10,5)+C(10,4))$
$(;All; 5-member; committiees;);-;(;committiees; with ;no; Republicans;+;committiees; with; one ;Republican ;)$
Explanation: There are $C(20,5)$ 5-member committiees which will be subtracted from sum of two numbers in order to achieve the answer; C(10,5) is number of 5-member committiees which include no Republicans (I've only counted Democrats and Independents ($8+2$)then found number of 5-member committiees ) and C(10,4) is number of 5-member committiees with one Republican (Like before I've excluded Republicans but counted 4-member committiees then added one Republican to each subset)
Book's Answer:
It's just said $C(10,2)C(18,3)$
combinatorics binomial-theorem
asked 1 hour ago
Abbas
897
3
The number of committiees with exactly one Republican is $10cdot C(10,4)$ (there are $10$ ways to choose this Republican). This does not match book's answer, though.
– ajotatxe
1 hour ago
I'm not sure about that because , there was no mention of difference between each Republican
– Abbas
1 hour ago
1
The book answer is wrong, it's an overcount. Use the same logic for 2 R, 1 D, 2-member committee with at least 1 R. Answer is 3, but book would give 2-choose-1 times 2-choose-1, which is 4.
– Gerry Myerson
1 hour ago
What text is this? Please notify the publisher of the error.
– N. F. Taussig
55 mins ago
Link to the text ufile.io/mnowz page 15 or 16
– Abbas
47 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
There are 10 Republicans, 8 Democrats and 2 Independent legislators eligible for committiee membership.
How many 5-member committiees exist which include at least two Republicans?
My Work :
$C(20,5)-(C(10,5)+C(10,4))$
$(;All; 5-member; committiees;);-;(;committiees; with ;no; Republicans;+;committiees; with; one ;Republican ;)$
Explanation: There are $C(20,5)$ 5-member committiees which will be subtracted from sum of two numbers in order to achieve the answer; C(10,5) is number of 5-member committiees which include no Republicans (I've only counted Democrats and Independents ($8+2$)then found number of 5-member committiees ) and C(10,4) is number of 5-member committiees with one Republican (Like before I've excluded Republicans but counted 4-member committiees then added one Republican to each subset)
Book's Answer:
It's just said $C(10,2)C(18,3)$
combinatorics binomial-theorem
asked 1 hour ago
Abbas
897
There are 10 Republicans, 8 Democrats and 2 Independent legislators eligible for committiee membership.
How many 5-member committiees exist which include at least two Republicans?
My Work :
$C(20,5)-(C(10,5)+C(10,4))$
$(;All; 5-member; committiees;);-;(;committiees; with ;no; Republicans;+;committiees; with; one ;Republican ;)$
Explanation: There are $C(20,5)$ 5-member committiees which will be subtracted from sum of two numbers in order to achieve the answer; C(10,5) is number of 5-member committiees which include no Republicans (I've only counted Democrats and Independents ($8+2$)then found number of 5-member committiees ) and C(10,4) is number of 5-member committiees with one Republican (Like before I've excluded Republicans but counted 4-member committiees then added one Republican to each subset)
Book's Answer:
It's just said $C(10,2)C(18,3)$
combinatorics binomial-theorem
combinatorics binomial-theorem
asked 1 hour ago
Abbas
897
asked 1 hour ago
Abbas
897
asked 1 hour ago
Abbas
897
asked 1 hour ago
Abbas
897
asked 1 hour ago
Abbas
897
897
3
The number of committiees with exactly one Republican is $10cdot C(10,4)$ (there are $10$ ways to choose this Republican). This does not match book's answer, though.
– ajotatxe
1 hour ago
I'm not sure about that because , there was no mention of difference between each Republican
– Abbas
1 hour ago
1
The book answer is wrong, it's an overcount. Use the same logic for 2 R, 1 D, 2-member committee with at least 1 R. Answer is 3, but book would give 2-choose-1 times 2-choose-1, which is 4.
– Gerry Myerson
1 hour ago
What text is this? Please notify the publisher of the error.
– N. F. Taussig
55 mins ago
Link to the text ufile.io/mnowz page 15 or 16
– Abbas
47 mins ago
add a comment |Â
3
The number of committiees with exactly one Republican is $10cdot C(10,4)$ (there are $10$ ways to choose this Republican). This does not match book's answer, though.
– ajotatxe
1 hour ago
I'm not sure about that because , there was no mention of difference between each Republican
– Abbas
1 hour ago
1
The book answer is wrong, it's an overcount. Use the same logic for 2 R, 1 D, 2-member committee with at least 1 R. Answer is 3, but book would give 2-choose-1 times 2-choose-1, which is 4.
– Gerry Myerson
1 hour ago
What text is this? Please notify the publisher of the error.
– N. F. Taussig
55 mins ago
Link to the text ufile.io/mnowz page 15 or 16
– Abbas
47 mins ago
3
3
The number of committiees with exactly one Republican is $10cdot C(10,4)$ (there are $10$ ways to choose this Republican). This does not match book's answer, though.
– ajotatxe
1 hour ago
The number of committiees with exactly one Republican is $10cdot C(10,4)$ (there are $10$ ways to choose this Republican). This does not match book's answer, though.
– ajotatxe
1 hour ago
I'm not sure about that because , there was no mention of difference between each Republican
– Abbas
1 hour ago
I'm not sure about that because , there was no mention of difference between each Republican
– Abbas
1 hour ago
1
1
The book answer is wrong, it's an overcount. Use the same logic for 2 R, 1 D, 2-member committee with at least 1 R. Answer is 3, but book would give 2-choose-1 times 2-choose-1, which is 4.
– Gerry Myerson
1 hour ago
The book answer is wrong, it's an overcount. Use the same logic for 2 R, 1 D, 2-member committee with at least 1 R. Answer is 3, but book would give 2-choose-1 times 2-choose-1, which is 4.
– Gerry Myerson
1 hour ago
What text is this? Please notify the publisher of the error.
– N. F. Taussig
55 mins ago
What text is this? Please notify the publisher of the error.
– N. F. Taussig
55 mins ago
Link to the text ufile.io/mnowz page 15 or 16
– Abbas
47 mins ago
Link to the text ufile.io/mnowz page 15 or 16
– Abbas
47 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
I think that the answer is
$$binom205-binom105-10binom104=13,152$$
and that book's answer is wrong. The idea behind book's answer seems to be the following:
First, we choose two Republicans. There are $C(10,2)$ ways to do it. Then we choose three any other members. There are $C(18,3)$ ways to do it. The problem is that then we are double (or even triple, etc) counting many possible choices. For an example, let's say that the Republicans are numbered from $1$ to $10$, and the other ones from $11$ to $20$. If we choose for first $1$ and $2$ and then $3$, $4$ and $12$ it is the same as choosing for first $3$ and $4$ and then $1$, $2$ and $12$.
Note (just in case you don't know): $binom ab$ is just another way (and more usual, if you ask me) to write $C(a,b)$.
answered 1 hour ago
ajotatxe
50.3k13185
Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least"
– Abbas
1 hour ago
@Abbas That would not explain the error. If there were exactly two Republicans, the count would be $binom102binom103$.
– N. F. Taussig
1 hour ago
@N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10?
– Abbas
50 mins ago
1
@Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $binom102binom103$.
– N. F. Taussig
47 mins ago
@N. F. Taussig Got it ! Thanks
– Abbas
44 mins ago
add a comment |Â
up vote
1
down vote
Republicans-$10$,Democrats-$8$,Independent legislator-$2$
There should be at least $2$ Republicans in the $5$ members committee.
So, there may be $2$ or $3$ or $4$ or $5$ republicans in the committee, as there are no restrictions on the numbers of the Democrats & Independent legislator in the committee-so we can treat these two groups as a single one of $8+2=10$ members.Call this new group the D-IL group.
The number of ways of forming such committee is
[($2$ Republicans & $3$ D-IL)or ($3$ Republicans & $2$ D-IL)or ($4$ Republicans & $1$ D-IL)or ($5$ Republicans & $0$ D-IL)]
=($2$ Republicans)$cdot$ ($3$ D-IL)+($3$ Republicans)$cdot$ ($2$ D-IL)+($4$ Republicans)$cdot$ ($1$ D-IL)+($5$ Republicans)$cdot$($0$ D-IL)
=$10 choose 2cdot 10 choose 3+10 choose 3cdot 10 choose 2+10 choose 4cdot 10 choose 1+10 choose 5cdot 10 choose 0=13152.$
answered 41 mins ago
Arjun Banerjee
1758
Thank you for answering !
– Abbas
34 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
I think that the answer is
$$binom205-binom105-10binom104=13,152$$
and that book's answer is wrong. The idea behind book's answer seems to be the following:
First, we choose two Republicans. There are $C(10,2)$ ways to do it. Then we choose three any other members. There are $C(18,3)$ ways to do it. The problem is that then we are double (or even triple, etc) counting many possible choices. For an example, let's say that the Republicans are numbered from $1$ to $10$, and the other ones from $11$ to $20$. If we choose for first $1$ and $2$ and then $3$, $4$ and $12$ it is the same as choosing for first $3$ and $4$ and then $1$, $2$ and $12$.
Note (just in case you don't know): $binom ab$ is just another way (and more usual, if you ask me) to write $C(a,b)$.
answered 1 hour ago
ajotatxe
50.3k13185
Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least"
– Abbas
1 hour ago
@Abbas That would not explain the error. If there were exactly two Republicans, the count would be $binom102binom103$.
– N. F. Taussig
1 hour ago
@N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10?
– Abbas
50 mins ago
1
@Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $binom102binom103$.
– N. F. Taussig
47 mins ago
@N. F. Taussig Got it ! Thanks
– Abbas
44 mins ago
add a comment |Â
up vote
4
down vote
accepted
I think that the answer is
$$binom205-binom105-10binom104=13,152$$
and that book's answer is wrong. The idea behind book's answer seems to be the following:
First, we choose two Republicans. There are $C(10,2)$ ways to do it. Then we choose three any other members. There are $C(18,3)$ ways to do it. The problem is that then we are double (or even triple, etc) counting many possible choices. For an example, let's say that the Republicans are numbered from $1$ to $10$, and the other ones from $11$ to $20$. If we choose for first $1$ and $2$ and then $3$, $4$ and $12$ it is the same as choosing for first $3$ and $4$ and then $1$, $2$ and $12$.
Note (just in case you don't know): $binom ab$ is just another way (and more usual, if you ask me) to write $C(a,b)$.
answered 1 hour ago
ajotatxe
50.3k13185
Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least"
– Abbas
1 hour ago
@Abbas That would not explain the error. If there were exactly two Republicans, the count would be $binom102binom103$.
– N. F. Taussig
1 hour ago
@N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10?
– Abbas
50 mins ago
1
@Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $binom102binom103$.
– N. F. Taussig
47 mins ago
@N. F. Taussig Got it ! Thanks
– Abbas
44 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
I think that the answer is
$$binom205-binom105-10binom104=13,152$$
and that book's answer is wrong. The idea behind book's answer seems to be the following:
First, we choose two Republicans. There are $C(10,2)$ ways to do it. Then we choose three any other members. There are $C(18,3)$ ways to do it. The problem is that then we are double (or even triple, etc) counting many possible choices. For an example, let's say that the Republicans are numbered from $1$ to $10$, and the other ones from $11$ to $20$. If we choose for first $1$ and $2$ and then $3$, $4$ and $12$ it is the same as choosing for first $3$ and $4$ and then $1$, $2$ and $12$.
Note (just in case you don't know): $binom ab$ is just another way (and more usual, if you ask me) to write $C(a,b)$.
answered 1 hour ago
ajotatxe
50.3k13185
I think that the answer is
$$binom205-binom105-10binom104=13,152$$
and that book's answer is wrong. The idea behind book's answer seems to be the following:
First, we choose two Republicans. There are $C(10,2)$ ways to do it. Then we choose three any other members. There are $C(18,3)$ ways to do it. The problem is that then we are double (or even triple, etc) counting many possible choices. For an example, let's say that the Republicans are numbered from $1$ to $10$, and the other ones from $11$ to $20$. If we choose for first $1$ and $2$ and then $3$, $4$ and $12$ it is the same as choosing for first $3$ and $4$ and then $1$, $2$ and $12$.
Note (just in case you don't know): $binom ab$ is just another way (and more usual, if you ask me) to write $C(a,b)$.
answered 1 hour ago
ajotatxe
50.3k13185
edited 1 hour ago
answered 1 hour ago
ajotatxe
50.3k13185
answered 1 hour ago
ajotatxe
50.3k13185
answered 1 hour ago
ajotatxe
50.3k13185
50.3k13185
Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least"
– Abbas
1 hour ago
@Abbas That would not explain the error. If there were exactly two Republicans, the count would be $binom102binom103$.
– N. F. Taussig
1 hour ago
@N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10?
– Abbas
50 mins ago
1
@Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $binom102binom103$.
– N. F. Taussig
47 mins ago
@N. F. Taussig Got it ! Thanks
– Abbas
44 mins ago
add a comment |Â
Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least"
– Abbas
1 hour ago
@Abbas That would not explain the error. If there were exactly two Republicans, the count would be $binom102binom103$.
– N. F. Taussig
1 hour ago
@N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10?
– Abbas
50 mins ago
1
@Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $binom102binom103$.
– N. F. Taussig
47 mins ago
@N. F. Taussig Got it ! Thanks
– Abbas
44 mins ago
Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least"
– Abbas
1 hour ago
Thank you for answering. I wonder if the book's approach counts all the ways because as you said it counts state of having two Republicans but it seems that it not paying attention to "at least"
– Abbas
1 hour ago
@Abbas That would not explain the error. If there were exactly two Republicans, the count would be $binom102binom103$.
– N. F. Taussig
1 hour ago
@Abbas That would not explain the error. If there were exactly two Republicans, the count would be $binom102binom103$.
– N. F. Taussig
1 hour ago
@N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10?
– Abbas
50 mins ago
@N. F. Taussig is $C(10,2)$ for removing the problem of 10 different Republicans ? Or putting it in another way , is it because of that 2 Republican can be any of that 10?
– Abbas
50 mins ago
1
1
@Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $binom102binom103$.
– N. F. Taussig
47 mins ago
@Abbas There are $10$ Republicans and $8 + 2 = 10$ non-Republicans who could serve on the committee. If there are exactly two Republicans on the five-member committee, there must be three non-Republicans. Thus, the number of ways exactly two Republicans can serve on the committee is the number of ways we can select two of the ten Republicans and three of the ten non-Republicans, which is $binom102binom103$.
– N. F. Taussig
47 mins ago
@N. F. Taussig Got it ! Thanks
– Abbas
44 mins ago
@N. F. Taussig Got it ! Thanks
– Abbas
44 mins ago
add a comment |Â
up vote
1
down vote
Republicans-$10$,Democrats-$8$,Independent legislator-$2$
There should be at least $2$ Republicans in the $5$ members committee.
So, there may be $2$ or $3$ or $4$ or $5$ republicans in the committee, as there are no restrictions on the numbers of the Democrats & Independent legislator in the committee-so we can treat these two groups as a single one of $8+2=10$ members.Call this new group the D-IL group.
The number of ways of forming such committee is
[($2$ Republicans & $3$ D-IL)or ($3$ Republicans & $2$ D-IL)or ($4$ Republicans & $1$ D-IL)or ($5$ Republicans & $0$ D-IL)]
=($2$ Republicans)$cdot$ ($3$ D-IL)+($3$ Republicans)$cdot$ ($2$ D-IL)+($4$ Republicans)$cdot$ ($1$ D-IL)+($5$ Republicans)$cdot$($0$ D-IL)
=$10 choose 2cdot 10 choose 3+10 choose 3cdot 10 choose 2+10 choose 4cdot 10 choose 1+10 choose 5cdot 10 choose 0=13152.$
answered 41 mins ago
Arjun Banerjee
1758
Thank you for answering !
– Abbas
34 mins ago
add a comment |Â
up vote
1
down vote
Republicans-$10$,Democrats-$8$,Independent legislator-$2$
There should be at least $2$ Republicans in the $5$ members committee.
So, there may be $2$ or $3$ or $4$ or $5$ republicans in the committee, as there are no restrictions on the numbers of the Democrats & Independent legislator in the committee-so we can treat these two groups as a single one of $8+2=10$ members.Call this new group the D-IL group.
The number of ways of forming such committee is
[($2$ Republicans & $3$ D-IL)or ($3$ Republicans & $2$ D-IL)or ($4$ Republicans & $1$ D-IL)or ($5$ Republicans & $0$ D-IL)]
=($2$ Republicans)$cdot$ ($3$ D-IL)+($3$ Republicans)$cdot$ ($2$ D-IL)+($4$ Republicans)$cdot$ ($1$ D-IL)+($5$ Republicans)$cdot$($0$ D-IL)
=$10 choose 2cdot 10 choose 3+10 choose 3cdot 10 choose 2+10 choose 4cdot 10 choose 1+10 choose 5cdot 10 choose 0=13152.$
answered 41 mins ago
Arjun Banerjee
1758
Thank you for answering !
– Abbas
34 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Republicans-$10$,Democrats-$8$,Independent legislator-$2$
There should be at least $2$ Republicans in the $5$ members committee.
So, there may be $2$ or $3$ or $4$ or $5$ republicans in the committee, as there are no restrictions on the numbers of the Democrats & Independent legislator in the committee-so we can treat these two groups as a single one of $8+2=10$ members.Call this new group the D-IL group.
The number of ways of forming such committee is
[($2$ Republicans & $3$ D-IL)or ($3$ Republicans & $2$ D-IL)or ($4$ Republicans & $1$ D-IL)or ($5$ Republicans & $0$ D-IL)]
=($2$ Republicans)$cdot$ ($3$ D-IL)+($3$ Republicans)$cdot$ ($2$ D-IL)+($4$ Republicans)$cdot$ ($1$ D-IL)+($5$ Republicans)$cdot$($0$ D-IL)
=$10 choose 2cdot 10 choose 3+10 choose 3cdot 10 choose 2+10 choose 4cdot 10 choose 1+10 choose 5cdot 10 choose 0=13152.$
answered 41 mins ago
Arjun Banerjee
1758
Republicans-$10$,Democrats-$8$,Independent legislator-$2$
There should be at least $2$ Republicans in the $5$ members committee.
So, there may be $2$ or $3$ or $4$ or $5$ republicans in the committee, as there are no restrictions on the numbers of the Democrats & Independent legislator in the committee-so we can treat these two groups as a single one of $8+2=10$ members.Call this new group the D-IL group.
The number of ways of forming such committee is
[($2$ Republicans & $3$ D-IL)or ($3$ Republicans & $2$ D-IL)or ($4$ Republicans & $1$ D-IL)or ($5$ Republicans & $0$ D-IL)]
=($2$ Republicans)$cdot$ ($3$ D-IL)+($3$ Republicans)$cdot$ ($2$ D-IL)+($4$ Republicans)$cdot$ ($1$ D-IL)+($5$ Republicans)$cdot$($0$ D-IL)
=$10 choose 2cdot 10 choose 3+10 choose 3cdot 10 choose 2+10 choose 4cdot 10 choose 1+10 choose 5cdot 10 choose 0=13152.$
answered 41 mins ago
Arjun Banerjee
1758
answered 41 mins ago
Arjun Banerjee
1758
answered 41 mins ago
Arjun Banerjee
1758
answered 41 mins ago
Arjun Banerjee
1758
1758
Thank you for answering !
– Abbas
34 mins ago
add a comment |Â
Thank you for answering !
– Abbas
34 mins ago
Thank you for answering !
– Abbas
34 mins ago
Thank you for answering !
– Abbas
34 mins ago
add a comment |Â
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3
The number of committiees with exactly one Republican is $10cdot C(10,4)$ (there are $10$ ways to choose this Republican). This does not match book's answer, though.
– ajotatxe
1 hour ago
I'm not sure about that because , there was no mention of difference between each Republican
– Abbas
1 hour ago
1
The book answer is wrong, it's an overcount. Use the same logic for 2 R, 1 D, 2-member committee with at least 1 R. Answer is 3, but book would give 2-choose-1 times 2-choose-1, which is 4.
– Gerry Myerson
1 hour ago
What text is this? Please notify the publisher of the error.
– N. F. Taussig
55 mins ago
Link to the text ufile.io/mnowz page 15 or 16
– Abbas
47 mins ago