How many possible ways there are to assign 30 students to 3 instructors, where each instructor is assigned...












7












$begingroup$


As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    7 hours ago
















7












$begingroup$


As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    7 hours ago














7












7








7


1



$begingroup$


As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics







share|cite|improve this question











$endgroup$




As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago







Kroyer

















asked 7 hours ago









KroyerKroyer

456




456












  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    7 hours ago


















  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    7 hours ago
















$begingroup$
For more information, look into multinomial coefficients.
$endgroup$
– Michael Seifert
7 hours ago




$begingroup$
For more information, look into multinomial coefficients.
$endgroup$
– Michael Seifert
7 hours ago










4 Answers
4






active

oldest

votes


















10












$begingroup$

Firs, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    6 hours ago





















8












$begingroup$

It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, now its's clear to me.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    7 hours ago










  • $begingroup$
    Couldn't choose a correct answer straight away, had a time limit but I think one other later answer explains it more clearly.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    This is a nice intuituve explanation for the final answer as presented above.
    $endgroup$
    – akozi
    51 mins ago



















6












$begingroup$

The answer on this is $$frac{30!}{10!10!10!}$$



First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



Now realize that every possible splitup will be counted $10!10!10!$ times.



So division by $10!10!10!$ will repair.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you too, It was so simple after all.
    $endgroup$
    – Kroyer
    7 hours ago










  • $begingroup$
    Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
    $endgroup$
    – user601297
    6 hours ago






  • 1




    $begingroup$
    No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
    $endgroup$
    – drhab
    5 hours ago



















0












$begingroup$

Would not an alternative answer be 30 x 29 x 28 = 24,360 ?



I'm no mathematician, but it seems to me that the solutions offered thus far all would apply only if one were to assume that it matters what sequence the students are chosen in.



To me, this would be logical only if students were to always remain in a set order or arrangement (such as a seating plan or line).



I believe the answers provided previously would be most suitable as answers to a question that explicitly states that order is to be considered, such as, "How many ways are there to arrange the 30 students in three lines, if each line has 10 students," or "How many ways can 30 students be arranged in 3 classrooms, each classroom having 10 students sitting at desks in fixed positions?"



If one does not assume a specific seating arrangement or that one is lining students up in a specific order, then one could interpret the original question as pertaining merely to the number of different groupings (regardless of sequence) that one could potentially create when dividing 30 students into 3 groups of 10.






share|cite|improve this answer








New contributor




Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Hi! Welcome to MSE! Could you please explain, why do you think 24360 is a correct answer. I would like to know the details of your proof.
    $endgroup$
    – Yanior Weg
    3 hours ago










  • $begingroup$
    This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
    $endgroup$
    – Brahadeesh
    3 hours ago










  • $begingroup$
    Thanks for your patience with my having opined in this matter as a total amateur. I just chanced upon this site and the question caught my attention. As I said, I'm no mathematician. Having now re-read the other posts I can now see that you had taken steps to account/correct for the many different sequences within groupings which do not affect the content alone, of the groupings. As drhab stated, "... every possible splitup will be counted 10!10!10! times. So division by 10!10!10! will repair."
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    I believe my main impetus for adding a comment before having thought it through properly, was the utter shock at reading that the correct answer is over 5 trillion different groupings. Far beyond what I had imagined! WOW.
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
    $endgroup$
    – darij grinberg
    8 secs ago











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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









10












$begingroup$

Firs, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    6 hours ago


















10












$begingroup$

Firs, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    6 hours ago
















10












10








10





$begingroup$

Firs, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






share|cite|improve this answer









$endgroup$



Firs, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 7 hours ago









Yanior WegYanior Weg

1,74311138




1,74311138








  • 2




    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    6 hours ago
















  • 2




    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    6 hours ago










2




2




$begingroup$
For reference, it's $5 550 996 791 340$.
$endgroup$
– Eric Duminil
6 hours ago






$begingroup$
For reference, it's $5 550 996 791 340$.
$endgroup$
– Eric Duminil
6 hours ago













8












$begingroup$

It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, now its's clear to me.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    7 hours ago










  • $begingroup$
    Couldn't choose a correct answer straight away, had a time limit but I think one other later answer explains it more clearly.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    This is a nice intuituve explanation for the final answer as presented above.
    $endgroup$
    – akozi
    51 mins ago
















8












$begingroup$

It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, now its's clear to me.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    7 hours ago










  • $begingroup$
    Couldn't choose a correct answer straight away, had a time limit but I think one other later answer explains it more clearly.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    This is a nice intuituve explanation for the final answer as presented above.
    $endgroup$
    – akozi
    51 mins ago














8












8








8





$begingroup$

It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






share|cite|improve this answer









$endgroup$



It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 7 hours ago









enedilenedil

771416




771416












  • $begingroup$
    Thank you for your answer, now its's clear to me.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    7 hours ago










  • $begingroup$
    Couldn't choose a correct answer straight away, had a time limit but I think one other later answer explains it more clearly.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    This is a nice intuituve explanation for the final answer as presented above.
    $endgroup$
    – akozi
    51 mins ago


















  • $begingroup$
    Thank you for your answer, now its's clear to me.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    7 hours ago










  • $begingroup$
    Couldn't choose a correct answer straight away, had a time limit but I think one other later answer explains it more clearly.
    $endgroup$
    – Kroyer
    7 hours ago












  • $begingroup$
    This is a nice intuituve explanation for the final answer as presented above.
    $endgroup$
    – akozi
    51 mins ago
















$begingroup$
Thank you for your answer, now its's clear to me.
$endgroup$
– Kroyer
7 hours ago






$begingroup$
Thank you for your answer, now its's clear to me.
$endgroup$
– Kroyer
7 hours ago














$begingroup$
@JanTugsbayar then you can mark this as correct answer
$endgroup$
– enedil
7 hours ago




$begingroup$
@JanTugsbayar then you can mark this as correct answer
$endgroup$
– enedil
7 hours ago












$begingroup$
Couldn't choose a correct answer straight away, had a time limit but I think one other later answer explains it more clearly.
$endgroup$
– Kroyer
7 hours ago






$begingroup$
Couldn't choose a correct answer straight away, had a time limit but I think one other later answer explains it more clearly.
$endgroup$
– Kroyer
7 hours ago














$begingroup$
This is a nice intuituve explanation for the final answer as presented above.
$endgroup$
– akozi
51 mins ago




$begingroup$
This is a nice intuituve explanation for the final answer as presented above.
$endgroup$
– akozi
51 mins ago











6












$begingroup$

The answer on this is $$frac{30!}{10!10!10!}$$



First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



Now realize that every possible splitup will be counted $10!10!10!$ times.



So division by $10!10!10!$ will repair.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you too, It was so simple after all.
    $endgroup$
    – Kroyer
    7 hours ago










  • $begingroup$
    Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
    $endgroup$
    – user601297
    6 hours ago






  • 1




    $begingroup$
    No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
    $endgroup$
    – drhab
    5 hours ago
















6












$begingroup$

The answer on this is $$frac{30!}{10!10!10!}$$



First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



Now realize that every possible splitup will be counted $10!10!10!$ times.



So division by $10!10!10!$ will repair.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you too, It was so simple after all.
    $endgroup$
    – Kroyer
    7 hours ago










  • $begingroup$
    Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
    $endgroup$
    – user601297
    6 hours ago






  • 1




    $begingroup$
    No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
    $endgroup$
    – drhab
    5 hours ago














6












6








6





$begingroup$

The answer on this is $$frac{30!}{10!10!10!}$$



First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



Now realize that every possible splitup will be counted $10!10!10!$ times.



So division by $10!10!10!$ will repair.






share|cite|improve this answer









$endgroup$



The answer on this is $$frac{30!}{10!10!10!}$$



First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



Now realize that every possible splitup will be counted $10!10!10!$ times.



So division by $10!10!10!$ will repair.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 7 hours ago









drhabdrhab

101k544130




101k544130












  • $begingroup$
    Thank you too, It was so simple after all.
    $endgroup$
    – Kroyer
    7 hours ago










  • $begingroup$
    Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
    $endgroup$
    – user601297
    6 hours ago






  • 1




    $begingroup$
    No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
    $endgroup$
    – drhab
    5 hours ago


















  • $begingroup$
    Thank you too, It was so simple after all.
    $endgroup$
    – Kroyer
    7 hours ago










  • $begingroup$
    Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
    $endgroup$
    – user601297
    6 hours ago






  • 1




    $begingroup$
    No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
    $endgroup$
    – drhab
    5 hours ago
















$begingroup$
Thank you too, It was so simple after all.
$endgroup$
– Kroyer
7 hours ago




$begingroup$
Thank you too, It was so simple after all.
$endgroup$
– Kroyer
7 hours ago












$begingroup$
Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
$endgroup$
– user601297
6 hours ago




$begingroup$
Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
$endgroup$
– user601297
6 hours ago




1




1




$begingroup$
No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
$endgroup$
– drhab
5 hours ago




$begingroup$
No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
$endgroup$
– drhab
5 hours ago











0












$begingroup$

Would not an alternative answer be 30 x 29 x 28 = 24,360 ?



I'm no mathematician, but it seems to me that the solutions offered thus far all would apply only if one were to assume that it matters what sequence the students are chosen in.



To me, this would be logical only if students were to always remain in a set order or arrangement (such as a seating plan or line).



I believe the answers provided previously would be most suitable as answers to a question that explicitly states that order is to be considered, such as, "How many ways are there to arrange the 30 students in three lines, if each line has 10 students," or "How many ways can 30 students be arranged in 3 classrooms, each classroom having 10 students sitting at desks in fixed positions?"



If one does not assume a specific seating arrangement or that one is lining students up in a specific order, then one could interpret the original question as pertaining merely to the number of different groupings (regardless of sequence) that one could potentially create when dividing 30 students into 3 groups of 10.






share|cite|improve this answer








New contributor




Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Hi! Welcome to MSE! Could you please explain, why do you think 24360 is a correct answer. I would like to know the details of your proof.
    $endgroup$
    – Yanior Weg
    3 hours ago










  • $begingroup$
    This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
    $endgroup$
    – Brahadeesh
    3 hours ago










  • $begingroup$
    Thanks for your patience with my having opined in this matter as a total amateur. I just chanced upon this site and the question caught my attention. As I said, I'm no mathematician. Having now re-read the other posts I can now see that you had taken steps to account/correct for the many different sequences within groupings which do not affect the content alone, of the groupings. As drhab stated, "... every possible splitup will be counted 10!10!10! times. So division by 10!10!10! will repair."
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    I believe my main impetus for adding a comment before having thought it through properly, was the utter shock at reading that the correct answer is over 5 trillion different groupings. Far beyond what I had imagined! WOW.
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
    $endgroup$
    – darij grinberg
    8 secs ago
















0












$begingroup$

Would not an alternative answer be 30 x 29 x 28 = 24,360 ?



I'm no mathematician, but it seems to me that the solutions offered thus far all would apply only if one were to assume that it matters what sequence the students are chosen in.



To me, this would be logical only if students were to always remain in a set order or arrangement (such as a seating plan or line).



I believe the answers provided previously would be most suitable as answers to a question that explicitly states that order is to be considered, such as, "How many ways are there to arrange the 30 students in three lines, if each line has 10 students," or "How many ways can 30 students be arranged in 3 classrooms, each classroom having 10 students sitting at desks in fixed positions?"



If one does not assume a specific seating arrangement or that one is lining students up in a specific order, then one could interpret the original question as pertaining merely to the number of different groupings (regardless of sequence) that one could potentially create when dividing 30 students into 3 groups of 10.






share|cite|improve this answer








New contributor




Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Hi! Welcome to MSE! Could you please explain, why do you think 24360 is a correct answer. I would like to know the details of your proof.
    $endgroup$
    – Yanior Weg
    3 hours ago










  • $begingroup$
    This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
    $endgroup$
    – Brahadeesh
    3 hours ago










  • $begingroup$
    Thanks for your patience with my having opined in this matter as a total amateur. I just chanced upon this site and the question caught my attention. As I said, I'm no mathematician. Having now re-read the other posts I can now see that you had taken steps to account/correct for the many different sequences within groupings which do not affect the content alone, of the groupings. As drhab stated, "... every possible splitup will be counted 10!10!10! times. So division by 10!10!10! will repair."
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    I believe my main impetus for adding a comment before having thought it through properly, was the utter shock at reading that the correct answer is over 5 trillion different groupings. Far beyond what I had imagined! WOW.
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
    $endgroup$
    – darij grinberg
    8 secs ago














0












0








0





$begingroup$

Would not an alternative answer be 30 x 29 x 28 = 24,360 ?



I'm no mathematician, but it seems to me that the solutions offered thus far all would apply only if one were to assume that it matters what sequence the students are chosen in.



To me, this would be logical only if students were to always remain in a set order or arrangement (such as a seating plan or line).



I believe the answers provided previously would be most suitable as answers to a question that explicitly states that order is to be considered, such as, "How many ways are there to arrange the 30 students in three lines, if each line has 10 students," or "How many ways can 30 students be arranged in 3 classrooms, each classroom having 10 students sitting at desks in fixed positions?"



If one does not assume a specific seating arrangement or that one is lining students up in a specific order, then one could interpret the original question as pertaining merely to the number of different groupings (regardless of sequence) that one could potentially create when dividing 30 students into 3 groups of 10.






share|cite|improve this answer








New contributor




Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



Would not an alternative answer be 30 x 29 x 28 = 24,360 ?



I'm no mathematician, but it seems to me that the solutions offered thus far all would apply only if one were to assume that it matters what sequence the students are chosen in.



To me, this would be logical only if students were to always remain in a set order or arrangement (such as a seating plan or line).



I believe the answers provided previously would be most suitable as answers to a question that explicitly states that order is to be considered, such as, "How many ways are there to arrange the 30 students in three lines, if each line has 10 students," or "How many ways can 30 students be arranged in 3 classrooms, each classroom having 10 students sitting at desks in fixed positions?"



If one does not assume a specific seating arrangement or that one is lining students up in a specific order, then one could interpret the original question as pertaining merely to the number of different groupings (regardless of sequence) that one could potentially create when dividing 30 students into 3 groups of 10.







share|cite|improve this answer








New contributor




Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 3 hours ago









Daniel DaleDaniel Dale

1




1




New contributor




Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Daniel Dale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Hi! Welcome to MSE! Could you please explain, why do you think 24360 is a correct answer. I would like to know the details of your proof.
    $endgroup$
    – Yanior Weg
    3 hours ago










  • $begingroup$
    This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
    $endgroup$
    – Brahadeesh
    3 hours ago










  • $begingroup$
    Thanks for your patience with my having opined in this matter as a total amateur. I just chanced upon this site and the question caught my attention. As I said, I'm no mathematician. Having now re-read the other posts I can now see that you had taken steps to account/correct for the many different sequences within groupings which do not affect the content alone, of the groupings. As drhab stated, "... every possible splitup will be counted 10!10!10! times. So division by 10!10!10! will repair."
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    I believe my main impetus for adding a comment before having thought it through properly, was the utter shock at reading that the correct answer is over 5 trillion different groupings. Far beyond what I had imagined! WOW.
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
    $endgroup$
    – darij grinberg
    8 secs ago


















  • $begingroup$
    Hi! Welcome to MSE! Could you please explain, why do you think 24360 is a correct answer. I would like to know the details of your proof.
    $endgroup$
    – Yanior Weg
    3 hours ago










  • $begingroup$
    This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
    $endgroup$
    – Brahadeesh
    3 hours ago










  • $begingroup$
    Thanks for your patience with my having opined in this matter as a total amateur. I just chanced upon this site and the question caught my attention. As I said, I'm no mathematician. Having now re-read the other posts I can now see that you had taken steps to account/correct for the many different sequences within groupings which do not affect the content alone, of the groupings. As drhab stated, "... every possible splitup will be counted 10!10!10! times. So division by 10!10!10! will repair."
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    I believe my main impetus for adding a comment before having thought it through properly, was the utter shock at reading that the correct answer is over 5 trillion different groupings. Far beyond what I had imagined! WOW.
    $endgroup$
    – Daniel Dale
    2 hours ago










  • $begingroup$
    If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
    $endgroup$
    – darij grinberg
    8 secs ago
















$begingroup$
Hi! Welcome to MSE! Could you please explain, why do you think 24360 is a correct answer. I would like to know the details of your proof.
$endgroup$
– Yanior Weg
3 hours ago




$begingroup$
Hi! Welcome to MSE! Could you please explain, why do you think 24360 is a correct answer. I would like to know the details of your proof.
$endgroup$
– Yanior Weg
3 hours ago












$begingroup$
This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
$endgroup$
– Brahadeesh
3 hours ago




$begingroup$
This does not really answer the question. If you have a different question, you can ask it by clicking Ask Question. You can also add a bounty to draw more attention to this question once you have enough reputation. - From Review
$endgroup$
– Brahadeesh
3 hours ago












$begingroup$
Thanks for your patience with my having opined in this matter as a total amateur. I just chanced upon this site and the question caught my attention. As I said, I'm no mathematician. Having now re-read the other posts I can now see that you had taken steps to account/correct for the many different sequences within groupings which do not affect the content alone, of the groupings. As drhab stated, "... every possible splitup will be counted 10!10!10! times. So division by 10!10!10! will repair."
$endgroup$
– Daniel Dale
2 hours ago




$begingroup$
Thanks for your patience with my having opined in this matter as a total amateur. I just chanced upon this site and the question caught my attention. As I said, I'm no mathematician. Having now re-read the other posts I can now see that you had taken steps to account/correct for the many different sequences within groupings which do not affect the content alone, of the groupings. As drhab stated, "... every possible splitup will be counted 10!10!10! times. So division by 10!10!10! will repair."
$endgroup$
– Daniel Dale
2 hours ago












$begingroup$
I believe my main impetus for adding a comment before having thought it through properly, was the utter shock at reading that the correct answer is over 5 trillion different groupings. Far beyond what I had imagined! WOW.
$endgroup$
– Daniel Dale
2 hours ago




$begingroup$
I believe my main impetus for adding a comment before having thought it through properly, was the utter shock at reading that the correct answer is over 5 trillion different groupings. Far beyond what I had imagined! WOW.
$endgroup$
– Daniel Dale
2 hours ago












$begingroup$
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
$endgroup$
– darij grinberg
8 secs ago




$begingroup$
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review
$endgroup$
– darij grinberg
8 secs ago


















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