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Why aren't these two solutions equivalent? Combinatorics problem
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Different answers from different formulations of combinatorics problemNice problem of combinatorics..Trouble Understanding this Combinatorics ProblemThe number of ways to distribute groups between professors. Combinatorics. Surjective problem?Seating 10 people in a circular tableCounting problem combinatorics with employees of a faculty.Committee Forming Combinatorics ProblemA combinatorics problem with positionsCard combinatorics - two answersWhere is my solution wrong for this combinatorics problem?
$begingroup$
I was given the following fact: there is a set $S$ of $11$ people, among which there are $4$ professors and $7$ students,
$S={p_1, p_2, p_3,p_4, s_1, s_2,...,s_7}$
We are requested to form from it a group of $5$ people, and we must have at least 3 professors.
I find that the two answers I will expose should be equivalent, but are not, and I can't figure out why.
Answer 1
The group of $5$ people must have at least $3$ professors. This means that three of the $5$ people will necessarily be a subset of $S_p$, the subset of $S$ containing only the professors. There are $binom{4}{3}$ subsets of $S_p$, and therefore I have $binom{4}{3}$ alternatives for the three professors that must be in the group.
Now that I've made sure this $3$ professors are in the group, I have $11-3=8$ people left to choose from. The remaining two persons of the group can either be professors or students, so I can pick any of those $8$. So for the two remaining places I have $binom{8}{2}$ alternatives. At last, I have $binom{4}{3} binom{8}{2} = 112$ ways of forming a group of $5$ people in which there will definitely be at least $3$ professors.
Answer 2
There are $4$ professors and, in my group of $5$ people, I must have at least $3$ of them. So I'll either have $3$ or $4$ professors.
If I have $3$ professors, I'll choose them from the $4$ professors, and fill the remaining two places with $2$ of $7$ students. This is $binom{4}{3} binom{7}{2}$.
If on the other hand I have $4$ professors, I'll have $binom{4}{4}$ alternatives for choosing them, and $binom{7}{1}$ ways of choosing a student for the remaining last place.
So at last there are $binom{4}{3}binom{7}{2}+binom{4}{4}binom{7}{1} = 91$ ways of making the group.
Doubt
As you can see, the answers are different. Answer $1$ says there are $112$ ways of making the group; answer two says $91$. However, both reasonings seem okay to me and I can't see why should they differ nor where. Perhaps someone can clear this up for me.
combinatorics discrete-mathematics problem-solving
$endgroup$
add a comment |
$begingroup$
I was given the following fact: there is a set $S$ of $11$ people, among which there are $4$ professors and $7$ students,
$S={p_1, p_2, p_3,p_4, s_1, s_2,...,s_7}$
We are requested to form from it a group of $5$ people, and we must have at least 3 professors.
I find that the two answers I will expose should be equivalent, but are not, and I can't figure out why.
Answer 1
The group of $5$ people must have at least $3$ professors. This means that three of the $5$ people will necessarily be a subset of $S_p$, the subset of $S$ containing only the professors. There are $binom{4}{3}$ subsets of $S_p$, and therefore I have $binom{4}{3}$ alternatives for the three professors that must be in the group.
Now that I've made sure this $3$ professors are in the group, I have $11-3=8$ people left to choose from. The remaining two persons of the group can either be professors or students, so I can pick any of those $8$. So for the two remaining places I have $binom{8}{2}$ alternatives. At last, I have $binom{4}{3} binom{8}{2} = 112$ ways of forming a group of $5$ people in which there will definitely be at least $3$ professors.
Answer 2
There are $4$ professors and, in my group of $5$ people, I must have at least $3$ of them. So I'll either have $3$ or $4$ professors.
If I have $3$ professors, I'll choose them from the $4$ professors, and fill the remaining two places with $2$ of $7$ students. This is $binom{4}{3} binom{7}{2}$.
If on the other hand I have $4$ professors, I'll have $binom{4}{4}$ alternatives for choosing them, and $binom{7}{1}$ ways of choosing a student for the remaining last place.
So at last there are $binom{4}{3}binom{7}{2}+binom{4}{4}binom{7}{1} = 91$ ways of making the group.
Doubt
As you can see, the answers are different. Answer $1$ says there are $112$ ways of making the group; answer two says $91$. However, both reasonings seem okay to me and I can't see why should they differ nor where. Perhaps someone can clear this up for me.
combinatorics discrete-mathematics problem-solving
$endgroup$
add a comment |
$begingroup$
I was given the following fact: there is a set $S$ of $11$ people, among which there are $4$ professors and $7$ students,
$S={p_1, p_2, p_3,p_4, s_1, s_2,...,s_7}$
We are requested to form from it a group of $5$ people, and we must have at least 3 professors.
I find that the two answers I will expose should be equivalent, but are not, and I can't figure out why.
Answer 1
The group of $5$ people must have at least $3$ professors. This means that three of the $5$ people will necessarily be a subset of $S_p$, the subset of $S$ containing only the professors. There are $binom{4}{3}$ subsets of $S_p$, and therefore I have $binom{4}{3}$ alternatives for the three professors that must be in the group.
Now that I've made sure this $3$ professors are in the group, I have $11-3=8$ people left to choose from. The remaining two persons of the group can either be professors or students, so I can pick any of those $8$. So for the two remaining places I have $binom{8}{2}$ alternatives. At last, I have $binom{4}{3} binom{8}{2} = 112$ ways of forming a group of $5$ people in which there will definitely be at least $3$ professors.
Answer 2
There are $4$ professors and, in my group of $5$ people, I must have at least $3$ of them. So I'll either have $3$ or $4$ professors.
If I have $3$ professors, I'll choose them from the $4$ professors, and fill the remaining two places with $2$ of $7$ students. This is $binom{4}{3} binom{7}{2}$.
If on the other hand I have $4$ professors, I'll have $binom{4}{4}$ alternatives for choosing them, and $binom{7}{1}$ ways of choosing a student for the remaining last place.
So at last there are $binom{4}{3}binom{7}{2}+binom{4}{4}binom{7}{1} = 91$ ways of making the group.
Doubt
As you can see, the answers are different. Answer $1$ says there are $112$ ways of making the group; answer two says $91$. However, both reasonings seem okay to me and I can't see why should they differ nor where. Perhaps someone can clear this up for me.
combinatorics discrete-mathematics problem-solving
$endgroup$
I was given the following fact: there is a set $S$ of $11$ people, among which there are $4$ professors and $7$ students,
$S={p_1, p_2, p_3,p_4, s_1, s_2,...,s_7}$
We are requested to form from it a group of $5$ people, and we must have at least 3 professors.
I find that the two answers I will expose should be equivalent, but are not, and I can't figure out why.
Answer 1
The group of $5$ people must have at least $3$ professors. This means that three of the $5$ people will necessarily be a subset of $S_p$, the subset of $S$ containing only the professors. There are $binom{4}{3}$ subsets of $S_p$, and therefore I have $binom{4}{3}$ alternatives for the three professors that must be in the group.
Now that I've made sure this $3$ professors are in the group, I have $11-3=8$ people left to choose from. The remaining two persons of the group can either be professors or students, so I can pick any of those $8$. So for the two remaining places I have $binom{8}{2}$ alternatives. At last, I have $binom{4}{3} binom{8}{2} = 112$ ways of forming a group of $5$ people in which there will definitely be at least $3$ professors.
Answer 2
There are $4$ professors and, in my group of $5$ people, I must have at least $3$ of them. So I'll either have $3$ or $4$ professors.
If I have $3$ professors, I'll choose them from the $4$ professors, and fill the remaining two places with $2$ of $7$ students. This is $binom{4}{3} binom{7}{2}$.
If on the other hand I have $4$ professors, I'll have $binom{4}{4}$ alternatives for choosing them, and $binom{7}{1}$ ways of choosing a student for the remaining last place.
So at last there are $binom{4}{3}binom{7}{2}+binom{4}{4}binom{7}{1} = 91$ ways of making the group.
Doubt
As you can see, the answers are different. Answer $1$ says there are $112$ ways of making the group; answer two says $91$. However, both reasonings seem okay to me and I can't see why should they differ nor where. Perhaps someone can clear this up for me.
combinatorics discrete-mathematics problem-solving
combinatorics discrete-mathematics problem-solving
edited 6 hours ago
N. F. Taussig
45.5k103358
45.5k103358
asked 7 hours ago
AngelusSilesiusAngelusSilesius
1137
1137
add a comment |
add a comment |
2 Answers
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$begingroup$
Your second solution is the correct one.
Your first solution is incorrect because you overcount the scenarios where a professor is picked in the second step.
The outcome where you pick the first three professors in the first step followed by the fourth professor in the second step: ${p_1,p_2,p_3},{p_4,s_1}$ is also counted where you picked the last three professors in the first step and the first professor in the second step: ${p_2,p_3,p_4},{p_1,s_1}$. These outcomes should be considered the same however since in both scenarios you have the same five people selected.
Be careful not to overcount things with multiplication principle. Objects selected in one step are treated differently than objects selected in a later step.
$endgroup$
$begingroup$
Of course! Now I understand. Thank you for answering my doubt.
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
The first answer is wrong. It overestimates the count by double-counting the four-professors solutions. This is because each can begin with three of the four in four different ways. Note that $$binom{4}{3}binom{7}{2}+4binom{4}{4}binom{7}{1}=112.$$Although "double counting" referred above to a fallacy, it's also the name of a valid, useful technique one should be happy to use.
$endgroup$
$begingroup$
This is right, thank you for taking the trouble and answering my doubt!
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
Your Answer
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2 Answers
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2 Answers
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active
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$begingroup$
Your second solution is the correct one.
Your first solution is incorrect because you overcount the scenarios where a professor is picked in the second step.
The outcome where you pick the first three professors in the first step followed by the fourth professor in the second step: ${p_1,p_2,p_3},{p_4,s_1}$ is also counted where you picked the last three professors in the first step and the first professor in the second step: ${p_2,p_3,p_4},{p_1,s_1}$. These outcomes should be considered the same however since in both scenarios you have the same five people selected.
Be careful not to overcount things with multiplication principle. Objects selected in one step are treated differently than objects selected in a later step.
$endgroup$
$begingroup$
Of course! Now I understand. Thank you for answering my doubt.
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
Your second solution is the correct one.
Your first solution is incorrect because you overcount the scenarios where a professor is picked in the second step.
The outcome where you pick the first three professors in the first step followed by the fourth professor in the second step: ${p_1,p_2,p_3},{p_4,s_1}$ is also counted where you picked the last three professors in the first step and the first professor in the second step: ${p_2,p_3,p_4},{p_1,s_1}$. These outcomes should be considered the same however since in both scenarios you have the same five people selected.
Be careful not to overcount things with multiplication principle. Objects selected in one step are treated differently than objects selected in a later step.
$endgroup$
$begingroup$
Of course! Now I understand. Thank you for answering my doubt.
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
Your second solution is the correct one.
Your first solution is incorrect because you overcount the scenarios where a professor is picked in the second step.
The outcome where you pick the first three professors in the first step followed by the fourth professor in the second step: ${p_1,p_2,p_3},{p_4,s_1}$ is also counted where you picked the last three professors in the first step and the first professor in the second step: ${p_2,p_3,p_4},{p_1,s_1}$. These outcomes should be considered the same however since in both scenarios you have the same five people selected.
Be careful not to overcount things with multiplication principle. Objects selected in one step are treated differently than objects selected in a later step.
$endgroup$
Your second solution is the correct one.
Your first solution is incorrect because you overcount the scenarios where a professor is picked in the second step.
The outcome where you pick the first three professors in the first step followed by the fourth professor in the second step: ${p_1,p_2,p_3},{p_4,s_1}$ is also counted where you picked the last three professors in the first step and the first professor in the second step: ${p_2,p_3,p_4},{p_1,s_1}$. These outcomes should be considered the same however since in both scenarios you have the same five people selected.
Be careful not to overcount things with multiplication principle. Objects selected in one step are treated differently than objects selected in a later step.
answered 7 hours ago
JMoravitzJMoravitz
49.5k44092
49.5k44092
$begingroup$
Of course! Now I understand. Thank you for answering my doubt.
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
Of course! Now I understand. Thank you for answering my doubt.
$endgroup$
– AngelusSilesius
6 hours ago
$begingroup$
Of course! Now I understand. Thank you for answering my doubt.
$endgroup$
– AngelusSilesius
6 hours ago
$begingroup$
Of course! Now I understand. Thank you for answering my doubt.
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
The first answer is wrong. It overestimates the count by double-counting the four-professors solutions. This is because each can begin with three of the four in four different ways. Note that $$binom{4}{3}binom{7}{2}+4binom{4}{4}binom{7}{1}=112.$$Although "double counting" referred above to a fallacy, it's also the name of a valid, useful technique one should be happy to use.
$endgroup$
$begingroup$
This is right, thank you for taking the trouble and answering my doubt!
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
The first answer is wrong. It overestimates the count by double-counting the four-professors solutions. This is because each can begin with three of the four in four different ways. Note that $$binom{4}{3}binom{7}{2}+4binom{4}{4}binom{7}{1}=112.$$Although "double counting" referred above to a fallacy, it's also the name of a valid, useful technique one should be happy to use.
$endgroup$
$begingroup$
This is right, thank you for taking the trouble and answering my doubt!
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
The first answer is wrong. It overestimates the count by double-counting the four-professors solutions. This is because each can begin with three of the four in four different ways. Note that $$binom{4}{3}binom{7}{2}+4binom{4}{4}binom{7}{1}=112.$$Although "double counting" referred above to a fallacy, it's also the name of a valid, useful technique one should be happy to use.
$endgroup$
The first answer is wrong. It overestimates the count by double-counting the four-professors solutions. This is because each can begin with three of the four in four different ways. Note that $$binom{4}{3}binom{7}{2}+4binom{4}{4}binom{7}{1}=112.$$Although "double counting" referred above to a fallacy, it's also the name of a valid, useful technique one should be happy to use.
edited 6 hours ago
answered 7 hours ago
J.G.J.G.
34.2k23252
34.2k23252
$begingroup$
This is right, thank you for taking the trouble and answering my doubt!
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
$begingroup$
This is right, thank you for taking the trouble and answering my doubt!
$endgroup$
– AngelusSilesius
6 hours ago
$begingroup$
This is right, thank you for taking the trouble and answering my doubt!
$endgroup$
– AngelusSilesius
6 hours ago
$begingroup$
This is right, thank you for taking the trouble and answering my doubt!
$endgroup$
– AngelusSilesius
6 hours ago
add a comment |
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