Mixing
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Prove for any positive integer $n$, $(4n)!$ is divisible by $2^{3n}cdot 3^n$
$begingroup$
Problem: Prove for any positive integer $n$, $(4n)!$ is divisible by $2^{3n}cdot 3^n$
Solution given by the professor: $$4! = 2^3cdot 3$$
$$(4!)^n = 2^{3n}cdot 3^n$$
$$frac{(4n)!}{(4!)^n}=frac{(4n)!}{2^{3n}cdot 3^n}$$
My question: The steps are pretty straightforward but I don't understand the last and most crucial step. For $frac{(4n)!}{2^{3n}cdot 3^n}$ to be an integer, we need $(4!)^n$ to divide $(4n)!$, is it a clear property of the factorial? How is it obvious?
combinatorics elementary-number-theory discrete-mathematics binomial-coefficients divisibility
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
add a comment |
$begingroup$
Problem: Prove for any positive integer $n$, $(4n)!$ is divisible by $2^{3n}cdot 3^n$
Solution given by the professor: $$4! = 2^3cdot 3$$
$$(4!)^n = 2^{3n}cdot 3^n$$
$$frac{(4n)!}{(4!)^n}=frac{(4n)!}{2^{3n}cdot 3^n}$$
My question: The steps are pretty straightforward but I don't understand the last and most crucial step. For $frac{(4n)!}{2^{3n}cdot 3^n}$ to be an integer, we need $(4!)^n$ to divide $(4n)!$, is it a clear property of the factorial? How is it obvious?
combinatorics elementary-number-theory discrete-mathematics binomial-coefficients divisibility
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
3
$begingroup$
Hint: if you write out $(4n)! = 4n times (4n-1) times (4n-2) times cdots times 2 times 1$, how many multiples of $4$ are there? How many multiples of $3$ and $2$?
$endgroup$
– Connor Harris
Jan 17 at 17:17
$begingroup$
@Wojowu Practice donning a kinder tone. Cheers!
$endgroup$
– jordan_glen
Jan 17 at 17:23
add a comment |
$begingroup$
Problem: Prove for any positive integer $n$, $(4n)!$ is divisible by $2^{3n}cdot 3^n$
Solution given by the professor: $$4! = 2^3cdot 3$$
$$(4!)^n = 2^{3n}cdot 3^n$$
$$frac{(4n)!}{(4!)^n}=frac{(4n)!}{2^{3n}cdot 3^n}$$
My question: The steps are pretty straightforward but I don't understand the last and most crucial step. For $frac{(4n)!}{2^{3n}cdot 3^n}$ to be an integer, we need $(4!)^n$ to divide $(4n)!$, is it a clear property of the factorial? How is it obvious?
combinatorics elementary-number-theory discrete-mathematics binomial-coefficients divisibility
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
$endgroup$
Problem: Prove for any positive integer $n$, $(4n)!$ is divisible by $2^{3n}cdot 3^n$
Solution given by the professor: $$4! = 2^3cdot 3$$
$$(4!)^n = 2^{3n}cdot 3^n$$
$$frac{(4n)!}{(4!)^n}=frac{(4n)!}{2^{3n}cdot 3^n}$$
My question: The steps are pretty straightforward but I don't understand the last and most crucial step. For $frac{(4n)!}{2^{3n}cdot 3^n}$ to be an integer, we need $(4!)^n$ to divide $(4n)!$, is it a clear property of the factorial? How is it obvious?
combinatorics elementary-number-theory discrete-mathematics binomial-coefficients divisibility
combinatorics elementary-number-theory discrete-mathematics binomial-coefficients divisibility
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
edited Jan 17 at 18:07
José Carlos Santos
163k22131234
163k22131234
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
asked Jan 17 at 17:14
NotAbelianGroupNotAbelianGroup
18211
18211
3
$begingroup$
Hint: if you write out $(4n)! = 4n times (4n-1) times (4n-2) times cdots times 2 times 1$, how many multiples of $4$ are there? How many multiples of $3$ and $2$?
$endgroup$
– Connor Harris
Jan 17 at 17:17
$begingroup$
@Wojowu Practice donning a kinder tone. Cheers!
$endgroup$
– jordan_glen
Jan 17 at 17:23
add a comment |
3
$begingroup$
Hint: if you write out $(4n)! = 4n times (4n-1) times (4n-2) times cdots times 2 times 1$, how many multiples of $4$ are there? How many multiples of $3$ and $2$?
$endgroup$
– Connor Harris
Jan 17 at 17:17
$begingroup$
@Wojowu Practice donning a kinder tone. Cheers!
$endgroup$
– jordan_glen
Jan 17 at 17:23
3
3
$begingroup$
Hint: if you write out $(4n)! = 4n times (4n-1) times (4n-2) times cdots times 2 times 1$, how many multiples of $4$ are there? How many multiples of $3$ and $2$?
$endgroup$
– Connor Harris
Jan 17 at 17:17
$begingroup$
Hint: if you write out $(4n)! = 4n times (4n-1) times (4n-2) times cdots times 2 times 1$, how many multiples of $4$ are there? How many multiples of $3$ and $2$?
$endgroup$
– Connor Harris
Jan 17 at 17:17
$begingroup$
@Wojowu Practice donning a kinder tone. Cheers!
$endgroup$
– jordan_glen
Jan 17 at 17:23
$begingroup$
@Wojowu Practice donning a kinder tone. Cheers!
$endgroup$
– jordan_glen
Jan 17 at 17:23
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
It is true that $(4!)^nmid(4n)!$, but it is not obvious (at least, not to me). You can use the fact that, for any natural number $m$, $4!mid m(m+1)(m+2)(m+3)$; after all$$frac{m(m+1)(m+2)(m+3)}{4!}=binom{m+3}4.$$So:
$4!mid1times2times3times4$;
$4!mid5times6times7times8$;- $vdots$
- $4!mid(4n-3)times(4n-2)times(4n-1)times(4n)$
and therefore $(4!)^nmid(4n)!$.
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
3
$begingroup$
Thank you for your answer, it is clearer now. In the notes the professor said to use the multinomial coefficient and I think I understand now how it relates : $$left( begin{array} { c } { n } \ { k _ { 1 } , k _ { 2 } , k _ { 3 } , ldots , k _ { m } } end{array} right) = left( begin{array} { c } { n } \ { vec { k } } end{array} right) = frac { n ! } { k _ { 1 } ! k _ { 2 } ! k _ { 3 } ! ldots k _ { m } ! }$$ Each one of my $k_i$ would be equal to $4$, and since I have $n$ of those $4$, the sum of all the $k_i$ equal to $4n$. Which is precisely the multinomial coefficient.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:40
1
$begingroup$
That is quite nice!
$endgroup$
– José Carlos Santos
Jan 17 at 17:42
add a comment |
$begingroup$
One can also use induction to show this. When $n =1$, the quotient is an integer$( = 1)$. Assume this is true for $n =k$ i.e., $(4k)!/(4!)^{k} in mathbb{Z}$. Then for $n =k+1$ we can see that $(4(k+1)!)/(4!)^{k+1} = {(4k+4)(4k+3)(4k+2)(4k+1)(4k!)}/4! times (4!)^{k}$
Then we need to prove that $(4k+1)(4k+2)(4k+3)(4k+4)/4!$ is an integer, which is indeed as pointed out in above post.
answered Jan 17 at 17:44
RickRick
868
$endgroup$
add a comment |
$begingroup$
Hint: The product of $k $ consecutive integers is divisible by $k! $.
For proving the above statement, consider the coefficients in binomial expansion.
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
$endgroup$
1
$begingroup$
Actually this is my next problem and this one I got it right: Let the consecutive numbers be $n , n - 1 , ldots , n - r + 1 .$ We have $$frac { n ( n - 1 ) cdots ( n - r + 1 ) } { r ! } = left( begin{array} { l } { n } \ { r } end{array} right)$$.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:35
$begingroup$
Cheers! {}{}{}{}{}
$endgroup$
– Thomas Shelby
Jan 17 at 18:03
add a comment |
$begingroup$
Hint: Every product of $m$ consecutive integers is divisible by $m!$. Let $m=4$.
answered Jan 17 at 17:22
ShaunShaun
9,300113684
$endgroup$
add a comment |
$begingroup$
Could proceed with induction:
n=1: $$ (4n)! = 4! = (4!)^n $$
n=2: $$ (4n)! = 8!= 8(7)(6)(5)(4)(3)(2)(1) mbox{ is divisible by } (4!)^n = 4^2(3^2)(2^2)(1) $$
Induction step: Assume that $(4!)^{n-1}$ divides $(4(n-1))!$
n: $$ frac{(4n)!}{(4!)^n} = frac{(4(n-1))! , times , mathbf{(4n-3)(4n-2)(4n-1)(4n)}}{(4!)^{n-1} , times , mathbf{4!}}$$ Notice that bold parts divide eachother trivially, and the non-bold parts divide by the induction hypothesis
answered Jan 17 at 17:33
NazimJNazimJ
36117
$endgroup$
add a comment |
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5 Answers
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active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is true that $(4!)^nmid(4n)!$, but it is not obvious (at least, not to me). You can use the fact that, for any natural number $m$, $4!mid m(m+1)(m+2)(m+3)$; after all$$frac{m(m+1)(m+2)(m+3)}{4!}=binom{m+3}4.$$So:
$4!mid1times2times3times4$;
$4!mid5times6times7times8$;- $vdots$
- $4!mid(4n-3)times(4n-2)times(4n-1)times(4n)$
and therefore $(4!)^nmid(4n)!$.
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
3
$begingroup$
Thank you for your answer, it is clearer now. In the notes the professor said to use the multinomial coefficient and I think I understand now how it relates : $$left( begin{array} { c } { n } \ { k _ { 1 } , k _ { 2 } , k _ { 3 } , ldots , k _ { m } } end{array} right) = left( begin{array} { c } { n } \ { vec { k } } end{array} right) = frac { n ! } { k _ { 1 } ! k _ { 2 } ! k _ { 3 } ! ldots k _ { m } ! }$$ Each one of my $k_i$ would be equal to $4$, and since I have $n$ of those $4$, the sum of all the $k_i$ equal to $4n$. Which is precisely the multinomial coefficient.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:40
1
$begingroup$
That is quite nice!
$endgroup$
– José Carlos Santos
Jan 17 at 17:42
add a comment |
$begingroup$
It is true that $(4!)^nmid(4n)!$, but it is not obvious (at least, not to me). You can use the fact that, for any natural number $m$, $4!mid m(m+1)(m+2)(m+3)$; after all$$frac{m(m+1)(m+2)(m+3)}{4!}=binom{m+3}4.$$So:
$4!mid1times2times3times4$;
$4!mid5times6times7times8$;- $vdots$
- $4!mid(4n-3)times(4n-2)times(4n-1)times(4n)$
and therefore $(4!)^nmid(4n)!$.
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
3
$begingroup$
Thank you for your answer, it is clearer now. In the notes the professor said to use the multinomial coefficient and I think I understand now how it relates : $$left( begin{array} { c } { n } \ { k _ { 1 } , k _ { 2 } , k _ { 3 } , ldots , k _ { m } } end{array} right) = left( begin{array} { c } { n } \ { vec { k } } end{array} right) = frac { n ! } { k _ { 1 } ! k _ { 2 } ! k _ { 3 } ! ldots k _ { m } ! }$$ Each one of my $k_i$ would be equal to $4$, and since I have $n$ of those $4$, the sum of all the $k_i$ equal to $4n$. Which is precisely the multinomial coefficient.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:40
1
$begingroup$
That is quite nice!
$endgroup$
– José Carlos Santos
Jan 17 at 17:42
add a comment |
$begingroup$
It is true that $(4!)^nmid(4n)!$, but it is not obvious (at least, not to me). You can use the fact that, for any natural number $m$, $4!mid m(m+1)(m+2)(m+3)$; after all$$frac{m(m+1)(m+2)(m+3)}{4!}=binom{m+3}4.$$So:
$4!mid1times2times3times4$;
$4!mid5times6times7times8$;- $vdots$
- $4!mid(4n-3)times(4n-2)times(4n-1)times(4n)$
and therefore $(4!)^nmid(4n)!$.
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
It is true that $(4!)^nmid(4n)!$, but it is not obvious (at least, not to me). You can use the fact that, for any natural number $m$, $4!mid m(m+1)(m+2)(m+3)$; after all$$frac{m(m+1)(m+2)(m+3)}{4!}=binom{m+3}4.$$So:
$4!mid1times2times3times4$;
$4!mid5times6times7times8$;- $vdots$
- $4!mid(4n-3)times(4n-2)times(4n-1)times(4n)$
and therefore $(4!)^nmid(4n)!$.
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 17:20
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
3
$begingroup$
Thank you for your answer, it is clearer now. In the notes the professor said to use the multinomial coefficient and I think I understand now how it relates : $$left( begin{array} { c } { n } \ { k _ { 1 } , k _ { 2 } , k _ { 3 } , ldots , k _ { m } } end{array} right) = left( begin{array} { c } { n } \ { vec { k } } end{array} right) = frac { n ! } { k _ { 1 } ! k _ { 2 } ! k _ { 3 } ! ldots k _ { m } ! }$$ Each one of my $k_i$ would be equal to $4$, and since I have $n$ of those $4$, the sum of all the $k_i$ equal to $4n$. Which is precisely the multinomial coefficient.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:40
1
$begingroup$
That is quite nice!
$endgroup$
– José Carlos Santos
Jan 17 at 17:42
add a comment |
3
$begingroup$
Thank you for your answer, it is clearer now. In the notes the professor said to use the multinomial coefficient and I think I understand now how it relates : $$left( begin{array} { c } { n } \ { k _ { 1 } , k _ { 2 } , k _ { 3 } , ldots , k _ { m } } end{array} right) = left( begin{array} { c } { n } \ { vec { k } } end{array} right) = frac { n ! } { k _ { 1 } ! k _ { 2 } ! k _ { 3 } ! ldots k _ { m } ! }$$ Each one of my $k_i$ would be equal to $4$, and since I have $n$ of those $4$, the sum of all the $k_i$ equal to $4n$. Which is precisely the multinomial coefficient.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:40
1
$begingroup$
That is quite nice!
$endgroup$
– José Carlos Santos
Jan 17 at 17:42
3
3
$begingroup$
Thank you for your answer, it is clearer now. In the notes the professor said to use the multinomial coefficient and I think I understand now how it relates : $$left( begin{array} { c } { n } \ { k _ { 1 } , k _ { 2 } , k _ { 3 } , ldots , k _ { m } } end{array} right) = left( begin{array} { c } { n } \ { vec { k } } end{array} right) = frac { n ! } { k _ { 1 } ! k _ { 2 } ! k _ { 3 } ! ldots k _ { m } ! }$$ Each one of my $k_i$ would be equal to $4$, and since I have $n$ of those $4$, the sum of all the $k_i$ equal to $4n$. Which is precisely the multinomial coefficient.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:40
$begingroup$
Thank you for your answer, it is clearer now. In the notes the professor said to use the multinomial coefficient and I think I understand now how it relates : $$left( begin{array} { c } { n } \ { k _ { 1 } , k _ { 2 } , k _ { 3 } , ldots , k _ { m } } end{array} right) = left( begin{array} { c } { n } \ { vec { k } } end{array} right) = frac { n ! } { k _ { 1 } ! k _ { 2 } ! k _ { 3 } ! ldots k _ { m } ! }$$ Each one of my $k_i$ would be equal to $4$, and since I have $n$ of those $4$, the sum of all the $k_i$ equal to $4n$. Which is precisely the multinomial coefficient.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:40
1
1
$begingroup$
That is quite nice!
$endgroup$
– José Carlos Santos
Jan 17 at 17:42
$begingroup$
That is quite nice!
$endgroup$
– José Carlos Santos
Jan 17 at 17:42
add a comment |
$begingroup$
One can also use induction to show this. When $n =1$, the quotient is an integer$( = 1)$. Assume this is true for $n =k$ i.e., $(4k)!/(4!)^{k} in mathbb{Z}$. Then for $n =k+1$ we can see that $(4(k+1)!)/(4!)^{k+1} = {(4k+4)(4k+3)(4k+2)(4k+1)(4k!)}/4! times (4!)^{k}$
Then we need to prove that $(4k+1)(4k+2)(4k+3)(4k+4)/4!$ is an integer, which is indeed as pointed out in above post.
answered Jan 17 at 17:44
RickRick
868
$endgroup$
add a comment |
$begingroup$
One can also use induction to show this. When $n =1$, the quotient is an integer$( = 1)$. Assume this is true for $n =k$ i.e., $(4k)!/(4!)^{k} in mathbb{Z}$. Then for $n =k+1$ we can see that $(4(k+1)!)/(4!)^{k+1} = {(4k+4)(4k+3)(4k+2)(4k+1)(4k!)}/4! times (4!)^{k}$
Then we need to prove that $(4k+1)(4k+2)(4k+3)(4k+4)/4!$ is an integer, which is indeed as pointed out in above post.
answered Jan 17 at 17:44
RickRick
868
$endgroup$
add a comment |
$begingroup$
One can also use induction to show this. When $n =1$, the quotient is an integer$( = 1)$. Assume this is true for $n =k$ i.e., $(4k)!/(4!)^{k} in mathbb{Z}$. Then for $n =k+1$ we can see that $(4(k+1)!)/(4!)^{k+1} = {(4k+4)(4k+3)(4k+2)(4k+1)(4k!)}/4! times (4!)^{k}$
Then we need to prove that $(4k+1)(4k+2)(4k+3)(4k+4)/4!$ is an integer, which is indeed as pointed out in above post.
answered Jan 17 at 17:44
RickRick
868
$endgroup$
One can also use induction to show this. When $n =1$, the quotient is an integer$( = 1)$. Assume this is true for $n =k$ i.e., $(4k)!/(4!)^{k} in mathbb{Z}$. Then for $n =k+1$ we can see that $(4(k+1)!)/(4!)^{k+1} = {(4k+4)(4k+3)(4k+2)(4k+1)(4k!)}/4! times (4!)^{k}$
Then we need to prove that $(4k+1)(4k+2)(4k+3)(4k+4)/4!$ is an integer, which is indeed as pointed out in above post.
answered Jan 17 at 17:44
RickRick
868
answered Jan 17 at 17:44
RickRick
868
answered Jan 17 at 17:44
RickRick
868
answered Jan 17 at 17:44
RickRick
868
868
add a comment |
add a comment |
$begingroup$
Hint: The product of $k $ consecutive integers is divisible by $k! $.
For proving the above statement, consider the coefficients in binomial expansion.
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
$endgroup$
1
$begingroup$
Actually this is my next problem and this one I got it right: Let the consecutive numbers be $n , n - 1 , ldots , n - r + 1 .$ We have $$frac { n ( n - 1 ) cdots ( n - r + 1 ) } { r ! } = left( begin{array} { l } { n } \ { r } end{array} right)$$.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:35
$begingroup$
Cheers! {}{}{}{}{}
$endgroup$
– Thomas Shelby
Jan 17 at 18:03
add a comment |
$begingroup$
Hint: The product of $k $ consecutive integers is divisible by $k! $.
For proving the above statement, consider the coefficients in binomial expansion.
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
$endgroup$
1
$begingroup$
Actually this is my next problem and this one I got it right: Let the consecutive numbers be $n , n - 1 , ldots , n - r + 1 .$ We have $$frac { n ( n - 1 ) cdots ( n - r + 1 ) } { r ! } = left( begin{array} { l } { n } \ { r } end{array} right)$$.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:35
$begingroup$
Cheers! {}{}{}{}{}
$endgroup$
– Thomas Shelby
Jan 17 at 18:03
add a comment |
$begingroup$
Hint: The product of $k $ consecutive integers is divisible by $k! $.
For proving the above statement, consider the coefficients in binomial expansion.
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
$endgroup$
Hint: The product of $k $ consecutive integers is divisible by $k! $.
For proving the above statement, consider the coefficients in binomial expansion.
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
answered Jan 17 at 17:19
Thomas ShelbyThomas Shelby
3,5642525
3,5642525
1
$begingroup$
Actually this is my next problem and this one I got it right: Let the consecutive numbers be $n , n - 1 , ldots , n - r + 1 .$ We have $$frac { n ( n - 1 ) cdots ( n - r + 1 ) } { r ! } = left( begin{array} { l } { n } \ { r } end{array} right)$$.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:35
$begingroup$
Cheers! {}{}{}{}{}
$endgroup$
– Thomas Shelby
Jan 17 at 18:03
add a comment |
1
$begingroup$
Actually this is my next problem and this one I got it right: Let the consecutive numbers be $n , n - 1 , ldots , n - r + 1 .$ We have $$frac { n ( n - 1 ) cdots ( n - r + 1 ) } { r ! } = left( begin{array} { l } { n } \ { r } end{array} right)$$.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:35
$begingroup$
Cheers! {}{}{}{}{}
$endgroup$
– Thomas Shelby
Jan 17 at 18:03
1
1
$begingroup$
Actually this is my next problem and this one I got it right: Let the consecutive numbers be $n , n - 1 , ldots , n - r + 1 .$ We have $$frac { n ( n - 1 ) cdots ( n - r + 1 ) } { r ! } = left( begin{array} { l } { n } \ { r } end{array} right)$$.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:35
$begingroup$
Actually this is my next problem and this one I got it right: Let the consecutive numbers be $n , n - 1 , ldots , n - r + 1 .$ We have $$frac { n ( n - 1 ) cdots ( n - r + 1 ) } { r ! } = left( begin{array} { l } { n } \ { r } end{array} right)$$.
$endgroup$
– NotAbelianGroup
Jan 17 at 17:35
$begingroup$
Cheers! {}{}{}{}{}
$endgroup$
– Thomas Shelby
Jan 17 at 18:03
$begingroup$
Cheers! {}{}{}{}{}
$endgroup$
– Thomas Shelby
Jan 17 at 18:03
add a comment |
$begingroup$
Hint: Every product of $m$ consecutive integers is divisible by $m!$. Let $m=4$.
answered Jan 17 at 17:22
ShaunShaun
9,300113684
$endgroup$
add a comment |
$begingroup$
Hint: Every product of $m$ consecutive integers is divisible by $m!$. Let $m=4$.
answered Jan 17 at 17:22
ShaunShaun
9,300113684
$endgroup$
add a comment |
$begingroup$
Hint: Every product of $m$ consecutive integers is divisible by $m!$. Let $m=4$.
answered Jan 17 at 17:22
ShaunShaun
9,300113684
$endgroup$
Hint: Every product of $m$ consecutive integers is divisible by $m!$. Let $m=4$.
answered Jan 17 at 17:22
ShaunShaun
9,300113684
answered Jan 17 at 17:22
ShaunShaun
9,300113684
answered Jan 17 at 17:22
ShaunShaun
9,300113684
answered Jan 17 at 17:22
ShaunShaun
9,300113684
9,300113684
add a comment |
add a comment |
$begingroup$
Could proceed with induction:
n=1: $$ (4n)! = 4! = (4!)^n $$
n=2: $$ (4n)! = 8!= 8(7)(6)(5)(4)(3)(2)(1) mbox{ is divisible by } (4!)^n = 4^2(3^2)(2^2)(1) $$
Induction step: Assume that $(4!)^{n-1}$ divides $(4(n-1))!$
n: $$ frac{(4n)!}{(4!)^n} = frac{(4(n-1))! , times , mathbf{(4n-3)(4n-2)(4n-1)(4n)}}{(4!)^{n-1} , times , mathbf{4!}}$$ Notice that bold parts divide eachother trivially, and the non-bold parts divide by the induction hypothesis
answered Jan 17 at 17:33
NazimJNazimJ
36117
$endgroup$
add a comment |
$begingroup$
Could proceed with induction:
n=1: $$ (4n)! = 4! = (4!)^n $$
n=2: $$ (4n)! = 8!= 8(7)(6)(5)(4)(3)(2)(1) mbox{ is divisible by } (4!)^n = 4^2(3^2)(2^2)(1) $$
Induction step: Assume that $(4!)^{n-1}$ divides $(4(n-1))!$
n: $$ frac{(4n)!}{(4!)^n} = frac{(4(n-1))! , times , mathbf{(4n-3)(4n-2)(4n-1)(4n)}}{(4!)^{n-1} , times , mathbf{4!}}$$ Notice that bold parts divide eachother trivially, and the non-bold parts divide by the induction hypothesis
answered Jan 17 at 17:33
NazimJNazimJ
36117
$endgroup$
add a comment |
$begingroup$
Could proceed with induction:
n=1: $$ (4n)! = 4! = (4!)^n $$
n=2: $$ (4n)! = 8!= 8(7)(6)(5)(4)(3)(2)(1) mbox{ is divisible by } (4!)^n = 4^2(3^2)(2^2)(1) $$
Induction step: Assume that $(4!)^{n-1}$ divides $(4(n-1))!$
n: $$ frac{(4n)!}{(4!)^n} = frac{(4(n-1))! , times , mathbf{(4n-3)(4n-2)(4n-1)(4n)}}{(4!)^{n-1} , times , mathbf{4!}}$$ Notice that bold parts divide eachother trivially, and the non-bold parts divide by the induction hypothesis
answered Jan 17 at 17:33
NazimJNazimJ
36117
$endgroup$
Could proceed with induction:
n=1: $$ (4n)! = 4! = (4!)^n $$
n=2: $$ (4n)! = 8!= 8(7)(6)(5)(4)(3)(2)(1) mbox{ is divisible by } (4!)^n = 4^2(3^2)(2^2)(1) $$
Induction step: Assume that $(4!)^{n-1}$ divides $(4(n-1))!$
n: $$ frac{(4n)!}{(4!)^n} = frac{(4(n-1))! , times , mathbf{(4n-3)(4n-2)(4n-1)(4n)}}{(4!)^{n-1} , times , mathbf{4!}}$$ Notice that bold parts divide eachother trivially, and the non-bold parts divide by the induction hypothesis
answered Jan 17 at 17:33
NazimJNazimJ
36117
answered Jan 17 at 17:33
NazimJNazimJ
36117
answered Jan 17 at 17:33
NazimJNazimJ
36117
answered Jan 17 at 17:33
NazimJNazimJ
36117
36117
add a comment |
add a comment |
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Hint: if you write out $(4n)! = 4n times (4n-1) times (4n-2) times cdots times 2 times 1$, how many multiples of $4$ are there? How many multiples of $3$ and $2$?
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– Connor Harris
Jan 17 at 17:17
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@Wojowu Practice donning a kinder tone. Cheers!
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– jordan_glen
Jan 17 at 17:23