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












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
    12 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
    12 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 3 hours ago









miracle173

7,33322247




7,33322247










asked 13 hours ago









KroyerKroyer

456




456












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


















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
















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




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










3 Answers
3






active

oldest

votes


















12












$begingroup$

First, 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











$endgroup$









  • 3




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





















11












$begingroup$

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
    12 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    12 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
    12 hours ago












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



















7












$begingroup$

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
    12 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
    12 hours ago






  • 2




    $begingroup$
    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
    10 hours ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









12












$begingroup$

First, 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











$endgroup$









  • 3




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


















12












$begingroup$

First, 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











$endgroup$









  • 3




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
















12












12








12





$begingroup$

First, 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











$endgroup$



First, 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








edited 4 hours ago









jvdhooft

5,25561641




5,25561641










answered 12 hours ago









Yanior WegYanior Weg

1,76311138




1,76311138








  • 3




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
















  • 3




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










3




3




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






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













11












$begingroup$

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
    12 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    12 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
    12 hours ago












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
















11












$begingroup$

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
    12 hours ago












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    12 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
    12 hours ago












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














11












11








11





$begingroup$

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 12 hours ago









enedilenedil

781418




781418












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












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    12 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
    12 hours ago












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


















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












  • $begingroup$
    @JanTugsbayar then you can mark this as correct answer
    $endgroup$
    – enedil
    12 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
    12 hours ago












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
















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






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














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




$begingroup$
@JanTugsbayar then you can mark this as correct answer
$endgroup$
– enedil
12 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
12 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
12 hours ago














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




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











7












$begingroup$

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
    12 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
    12 hours ago






  • 2




    $begingroup$
    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
    10 hours ago
















7












$begingroup$

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
    12 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
    12 hours ago






  • 2




    $begingroup$
    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
    10 hours ago














7












7








7





$begingroup$

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 12 hours ago









drhabdrhab

101k544130




101k544130












  • $begingroup$
    Thank you too, It was so simple after all.
    $endgroup$
    – Kroyer
    12 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
    12 hours ago






  • 2




    $begingroup$
    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
    10 hours ago


















  • $begingroup$
    Thank you too, It was so simple after all.
    $endgroup$
    – Kroyer
    12 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
    12 hours ago






  • 2




    $begingroup$
    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
    10 hours ago
















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




$begingroup$
Thank you too, It was so simple after all.
$endgroup$
– Kroyer
12 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
12 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
12 hours ago




2




2




$begingroup$
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
10 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
10 hours ago


















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RAC Tourist Trophy

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Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
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