Mixing
Equations with factorials
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
Solve the following equation :
$$4 (x+1)!= x! (2 x-6)!$$
My turn :$$24(x+1)!=6 x! (2x-6)!$$
$$4!(x+1)!=3!x!(2x-6)!$$
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
I tried to get a formula of a permutation in both sides but i could not do it
combinatorics
asked Sep 7 at 3:20
Hussien Mohamed
726112
add a comment |Â
up vote
9
down vote
favorite
Solve the following equation :
$$4 (x+1)!= x! (2 x-6)!$$
My turn :$$24(x+1)!=6 x! (2x-6)!$$
$$4!(x+1)!=3!x!(2x-6)!$$
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
I tried to get a formula of a permutation in both sides but i could not do it
combinatorics
asked Sep 7 at 3:20
Hussien Mohamed
726112
2
$4(x+1)!$ on LHS means the $4[(x+1)!]$ or $[4(x+1)]!$?
– Yuta
Sep 7 at 3:25
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Solve the following equation :
$$4 (x+1)!= x! (2 x-6)!$$
My turn :$$24(x+1)!=6 x! (2x-6)!$$
$$4!(x+1)!=3!x!(2x-6)!$$
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
I tried to get a formula of a permutation in both sides but i could not do it
combinatorics
asked Sep 7 at 3:20
Hussien Mohamed
726112
Solve the following equation :
$$4 (x+1)!= x! (2 x-6)!$$
My turn :$$24(x+1)!=6 x! (2x-6)!$$
$$4!(x+1)!=3!x!(2x-6)!$$
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
I tried to get a formula of a permutation in both sides but i could not do it
combinatorics
asked Sep 7 at 3:20
Hussien Mohamed
726112
asked Sep 7 at 3:20
Hussien Mohamed
726112
asked Sep 7 at 3:20
Hussien Mohamed
726112
asked Sep 7 at 3:20
Hussien Mohamed
726112
726112
2
$4(x+1)!$ on LHS means the $4[(x+1)!]$ or $[4(x+1)]!$?
– Yuta
Sep 7 at 3:25
add a comment |Â
2
$4(x+1)!$ on LHS means the $4[(x+1)!]$ or $[4(x+1)]!$?
– Yuta
Sep 7 at 3:25
2
2
$4(x+1)!$ on LHS means the $4[(x+1)!]$ or $[4(x+1)]!$?
– Yuta
Sep 7 at 3:25
$4(x+1)!$ on LHS means the $4[(x+1)!]$ or $[4(x+1)]!$?
– Yuta
Sep 7 at 3:25
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
10
down vote
accepted
This
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
is rewritten as
$$frac(x+1)x!3!x!=frac(2x-6)!4!$$
or even more
$$frac(x+1)x!3!x!=frac(2x-6)!4 cdot 3!$$
will give you
beginequation
x+1
=
frac14
(2x - 6)!
endequation
which is valid for $x = 5$.
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
4
You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique.
– Andrei
Sep 7 at 3:27
could you please edit it @Andrei ?
– Ahmad Bazzi
Sep 7 at 3:28
add a comment |Â
up vote
13
down vote
Bringing $x!$ to the left hand side gives $frac4(x+1)!x! = (2x - 6)! implies 4x + 4 = (2x-6)!$. Note that $(2x-6)!$ is a multiple of $4$, so this forces $2x - 6 geq 4$, as no other smaller factorial is a multiple of $4$.
Finally, note that $2x - 6 = 4 implies x = 5$ works, so we have one solution.There is no other solution, because if $x > 5$ and $4x + 4 < (2x - 6)!$, then $$4(x+1) + 4 = 4x + 8 < (4x+4)(2x-5)(2x-4) < (2(x+1) - 6)!$$
where the middle inequality is obvious from the fact that $frac4x+84x+4 = 1+frac44x+4$ which is smaller than $2$ if $x >1$ and the product of $2x - 5$ and $2x - 4$ is of course at least $2$ if $x > 3$. Consequently, by induction there is no solution for $x > 5$ (base case easily verified) and clearly , for $2x - 6 geq 0$ we must have $x geq 3$ and one sees that $x = 3,4$ do not work.
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
This
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
is rewritten as
$$frac(x+1)x!3!x!=frac(2x-6)!4!$$
or even more
$$frac(x+1)x!3!x!=frac(2x-6)!4 cdot 3!$$
will give you
beginequation
x+1
=
frac14
(2x - 6)!
endequation
which is valid for $x = 5$.
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
4
You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique.
– Andrei
Sep 7 at 3:27
could you please edit it @Andrei ?
– Ahmad Bazzi
Sep 7 at 3:28
add a comment |Â
up vote
10
down vote
accepted
This
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
is rewritten as
$$frac(x+1)x!3!x!=frac(2x-6)!4!$$
or even more
$$frac(x+1)x!3!x!=frac(2x-6)!4 cdot 3!$$
will give you
beginequation
x+1
=
frac14
(2x - 6)!
endequation
which is valid for $x = 5$.
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
4
You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique.
– Andrei
Sep 7 at 3:27
could you please edit it @Andrei ?
– Ahmad Bazzi
Sep 7 at 3:28
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
This
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
is rewritten as
$$frac(x+1)x!3!x!=frac(2x-6)!4!$$
or even more
$$frac(x+1)x!3!x!=frac(2x-6)!4 cdot 3!$$
will give you
beginequation
x+1
=
frac14
(2x - 6)!
endequation
which is valid for $x = 5$.
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
This
$$frac(x+1)!3!x!=frac(2x-6)!4!$$
is rewritten as
$$frac(x+1)x!3!x!=frac(2x-6)!4!$$
or even more
$$frac(x+1)x!3!x!=frac(2x-6)!4 cdot 3!$$
will give you
beginequation
x+1
=
frac14
(2x - 6)!
endequation
which is valid for $x = 5$.
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
edited 4 hours ago
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
answered Sep 7 at 3:25
Ahmad Bazzi
4,6701623
4,6701623
4
You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique.
– Andrei
Sep 7 at 3:27
could you please edit it @Andrei ?
– Ahmad Bazzi
Sep 7 at 3:28
add a comment |Â
4
You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique.
– Andrei
Sep 7 at 3:27
could you please edit it @Andrei ?
– Ahmad Bazzi
Sep 7 at 3:28
4
4
You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique.
– Andrei
Sep 7 at 3:27
You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique.
– Andrei
Sep 7 at 3:27
could you please edit it @Andrei ?
– Ahmad Bazzi
Sep 7 at 3:28
could you please edit it @Andrei ?
– Ahmad Bazzi
Sep 7 at 3:28
add a comment |Â
up vote
13
down vote
Bringing $x!$ to the left hand side gives $frac4(x+1)!x! = (2x - 6)! implies 4x + 4 = (2x-6)!$. Note that $(2x-6)!$ is a multiple of $4$, so this forces $2x - 6 geq 4$, as no other smaller factorial is a multiple of $4$.
Finally, note that $2x - 6 = 4 implies x = 5$ works, so we have one solution.There is no other solution, because if $x > 5$ and $4x + 4 < (2x - 6)!$, then $$4(x+1) + 4 = 4x + 8 < (4x+4)(2x-5)(2x-4) < (2(x+1) - 6)!$$
where the middle inequality is obvious from the fact that $frac4x+84x+4 = 1+frac44x+4$ which is smaller than $2$ if $x >1$ and the product of $2x - 5$ and $2x - 4$ is of course at least $2$ if $x > 3$. Consequently, by induction there is no solution for $x > 5$ (base case easily verified) and clearly , for $2x - 6 geq 0$ we must have $x geq 3$ and one sees that $x = 3,4$ do not work.
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
add a comment |Â
up vote
13
down vote
Bringing $x!$ to the left hand side gives $frac4(x+1)!x! = (2x - 6)! implies 4x + 4 = (2x-6)!$. Note that $(2x-6)!$ is a multiple of $4$, so this forces $2x - 6 geq 4$, as no other smaller factorial is a multiple of $4$.
Finally, note that $2x - 6 = 4 implies x = 5$ works, so we have one solution.There is no other solution, because if $x > 5$ and $4x + 4 < (2x - 6)!$, then $$4(x+1) + 4 = 4x + 8 < (4x+4)(2x-5)(2x-4) < (2(x+1) - 6)!$$
where the middle inequality is obvious from the fact that $frac4x+84x+4 = 1+frac44x+4$ which is smaller than $2$ if $x >1$ and the product of $2x - 5$ and $2x - 4$ is of course at least $2$ if $x > 3$. Consequently, by induction there is no solution for $x > 5$ (base case easily verified) and clearly , for $2x - 6 geq 0$ we must have $x geq 3$ and one sees that $x = 3,4$ do not work.
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
add a comment |Â
up vote
13
down vote
up vote
13
down vote
Bringing $x!$ to the left hand side gives $frac4(x+1)!x! = (2x - 6)! implies 4x + 4 = (2x-6)!$. Note that $(2x-6)!$ is a multiple of $4$, so this forces $2x - 6 geq 4$, as no other smaller factorial is a multiple of $4$.
Finally, note that $2x - 6 = 4 implies x = 5$ works, so we have one solution.There is no other solution, because if $x > 5$ and $4x + 4 < (2x - 6)!$, then $$4(x+1) + 4 = 4x + 8 < (4x+4)(2x-5)(2x-4) < (2(x+1) - 6)!$$
where the middle inequality is obvious from the fact that $frac4x+84x+4 = 1+frac44x+4$ which is smaller than $2$ if $x >1$ and the product of $2x - 5$ and $2x - 4$ is of course at least $2$ if $x > 3$. Consequently, by induction there is no solution for $x > 5$ (base case easily verified) and clearly , for $2x - 6 geq 0$ we must have $x geq 3$ and one sees that $x = 3,4$ do not work.
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
Bringing $x!$ to the left hand side gives $frac4(x+1)!x! = (2x - 6)! implies 4x + 4 = (2x-6)!$. Note that $(2x-6)!$ is a multiple of $4$, so this forces $2x - 6 geq 4$, as no other smaller factorial is a multiple of $4$.
Finally, note that $2x - 6 = 4 implies x = 5$ works, so we have one solution.There is no other solution, because if $x > 5$ and $4x + 4 < (2x - 6)!$, then $$4(x+1) + 4 = 4x + 8 < (4x+4)(2x-5)(2x-4) < (2(x+1) - 6)!$$
where the middle inequality is obvious from the fact that $frac4x+84x+4 = 1+frac44x+4$ which is smaller than $2$ if $x >1$ and the product of $2x - 5$ and $2x - 4$ is of course at least $2$ if $x > 3$. Consequently, by induction there is no solution for $x > 5$ (base case easily verified) and clearly , for $2x - 6 geq 0$ we must have $x geq 3$ and one sees that $x = 3,4$ do not work.
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
edited Sep 7 at 5:53
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
answered Sep 7 at 3:30
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
33.5k32870
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908234%2fequations-with-factorials%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
$4(x+1)!$ on LHS means the $4[(x+1)!]$ or $[4(x+1)]!$?
– Yuta
Sep 7 at 3:25