Quotient of factorials The 2019 Stack Overflow Developer Survey Results Are InThe product of $n$ consecutive integers is divisible by $n$ factoriallower bound for factorials productHow to evaluate factorials greater than $69!$Summing of factorials to produce perfect cubesFactorials…How do they do it?Show $frac(2n)!n!cdot 2^n$ is an integer for $n$ greater than or equal to $0$Solving equations with factorials?Is there an equivalent to the Bertrand's postulate between factorials and primorials?On the relationship between exponents and factorialsLongest sequence of consecutive integers which are not coprime with $n!$
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Quotient of factorials
The 2019 Stack Overflow Developer Survey Results Are InThe product of $n$ consecutive integers is divisible by $n$ factoriallower bound for factorials productHow to evaluate factorials greater than $69!$Summing of factorials to produce perfect cubesFactorials…How do they do it?Show $frac(2n)!n!cdot 2^n$ is an integer for $n$ greater than or equal to $0$Solving equations with factorials?Is there an equivalent to the Bertrand's postulate between factorials and primorials?On the relationship between exponents and factorialsLongest sequence of consecutive integers which are not coprime with $n!$
$begingroup$
Prove that $$(n^2)!over(n!)^n+1$$ is an integer, where $n$ is a natural number greater than $5$.
I know how the product of $r$ consecutive numbers is divisible by $r!$ Could we use it here? If so how, if not please help with any other suitable method.
factorial
edited Mar 24 at 8:05
MarianD
2,2611618
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
$endgroup$
add a comment |
$begingroup$
Prove that $$(n^2)!over(n!)^n+1$$ is an integer, where $n$ is a natural number greater than $5$.
I know how the product of $r$ consecutive numbers is divisible by $r!$ Could we use it here? If so how, if not please help with any other suitable method.
factorial
edited Mar 24 at 8:05
MarianD
2,2611618
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
$endgroup$
1
$begingroup$
This is the number of ways in which you can divide a group of $n^2$ objects into $n$ groups of $n$ objects each when there is no ordering within the groups or between the groups.
$endgroup$
– WimC
Mar 24 at 6:41
$begingroup$
Thanks I understood
$endgroup$
– Tarun Elango
Mar 24 at 6:44
add a comment |
$begingroup$
Prove that $$(n^2)!over(n!)^n+1$$ is an integer, where $n$ is a natural number greater than $5$.
I know how the product of $r$ consecutive numbers is divisible by $r!$ Could we use it here? If so how, if not please help with any other suitable method.
factorial
edited Mar 24 at 8:05
MarianD
2,2611618
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
$endgroup$
Prove that $$(n^2)!over(n!)^n+1$$ is an integer, where $n$ is a natural number greater than $5$.
I know how the product of $r$ consecutive numbers is divisible by $r!$ Could we use it here? If so how, if not please help with any other suitable method.
factorial
factorial
edited Mar 24 at 8:05
MarianD
2,2611618
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
edited Mar 24 at 8:05
MarianD
2,2611618
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
edited Mar 24 at 8:05
MarianD
2,2611618
edited Mar 24 at 8:05
MarianD
2,2611618
edited Mar 24 at 8:05
MarianD
2,2611618
2,2611618
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
asked Mar 24 at 6:35
Tarun ElangoTarun Elango
93
93
1
$begingroup$
This is the number of ways in which you can divide a group of $n^2$ objects into $n$ groups of $n$ objects each when there is no ordering within the groups or between the groups.
$endgroup$
– WimC
Mar 24 at 6:41
$begingroup$
Thanks I understood
$endgroup$
– Tarun Elango
Mar 24 at 6:44
add a comment |
1
$begingroup$
This is the number of ways in which you can divide a group of $n^2$ objects into $n$ groups of $n$ objects each when there is no ordering within the groups or between the groups.
$endgroup$
– WimC
Mar 24 at 6:41
$begingroup$
Thanks I understood
$endgroup$
– Tarun Elango
Mar 24 at 6:44
1
1
$begingroup$
This is the number of ways in which you can divide a group of $n^2$ objects into $n$ groups of $n$ objects each when there is no ordering within the groups or between the groups.
$endgroup$
– WimC
Mar 24 at 6:41
$begingroup$
This is the number of ways in which you can divide a group of $n^2$ objects into $n$ groups of $n$ objects each when there is no ordering within the groups or between the groups.
$endgroup$
– WimC
Mar 24 at 6:41
$begingroup$
Thanks I understood
$endgroup$
– Tarun Elango
Mar 24 at 6:44
$begingroup$
Thanks I understood
$endgroup$
– Tarun Elango
Mar 24 at 6:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's $$frac1n!binomn^2n,n,...,n,$$ which is a natural number for all natural $n$ because there are $n!$ permutations exactly of $(n,n,...,n).$
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand.
$endgroup$
– Tarun Elango
Mar 24 at 6:52
$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Mar 24 at 6:53
add a comment |
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1 Answer
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oldest
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1 Answer
1
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oldest
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active
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votes
$begingroup$
It's $$frac1n!binomn^2n,n,...,n,$$ which is a natural number for all natural $n$ because there are $n!$ permutations exactly of $(n,n,...,n).$
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand.
$endgroup$
– Tarun Elango
Mar 24 at 6:52
$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Mar 24 at 6:53
add a comment |
$begingroup$
It's $$frac1n!binomn^2n,n,...,n,$$ which is a natural number for all natural $n$ because there are $n!$ permutations exactly of $(n,n,...,n).$
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand.
$endgroup$
– Tarun Elango
Mar 24 at 6:52
$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Mar 24 at 6:53
add a comment |
$begingroup$
It's $$frac1n!binomn^2n,n,...,n,$$ which is a natural number for all natural $n$ because there are $n!$ permutations exactly of $(n,n,...,n).$
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
It's $$frac1n!binomn^2n,n,...,n,$$ which is a natural number for all natural $n$ because there are $n!$ permutations exactly of $(n,n,...,n).$
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 24 at 6:50
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
$begingroup$
Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand.
$endgroup$
– Tarun Elango
Mar 24 at 6:52
$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Mar 24 at 6:53
add a comment |
$begingroup$
Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand.
$endgroup$
– Tarun Elango
Mar 24 at 6:52
$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Mar 24 at 6:53
$begingroup$
Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand.
$endgroup$
– Tarun Elango
Mar 24 at 6:52
$begingroup$
Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand.
$endgroup$
– Tarun Elango
Mar 24 at 6:52
$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Mar 24 at 6:53
$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Mar 24 at 6:53
add a comment |
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1
$begingroup$
This is the number of ways in which you can divide a group of $n^2$ objects into $n$ groups of $n$ objects each when there is no ordering within the groups or between the groups.
$endgroup$
– WimC
Mar 24 at 6:41
$begingroup$
Thanks I understood
$endgroup$
– Tarun Elango
Mar 24 at 6:44