Solving for $r$ in ${12choose{r}}=924$
Multi tool use
$begingroup$
I can solve the equation $_{12}C_r=924$ fairly easily by guess and test because there are so few possible $r$ values, but is there a clean way to solve an equation of this format algebraically? I can't seem to simplify the combination expression without using a large number of cases.
combinations factorial
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
$endgroup$
add a comment |
$begingroup$
I can solve the equation $_{12}C_r=924$ fairly easily by guess and test because there are so few possible $r$ values, but is there a clean way to solve an equation of this format algebraically? I can't seem to simplify the combination expression without using a large number of cases.
combinations factorial
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
$endgroup$
add a comment |
$begingroup$
I can solve the equation $_{12}C_r=924$ fairly easily by guess and test because there are so few possible $r$ values, but is there a clean way to solve an equation of this format algebraically? I can't seem to simplify the combination expression without using a large number of cases.
combinations factorial
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
$endgroup$
I can solve the equation $_{12}C_r=924$ fairly easily by guess and test because there are so few possible $r$ values, but is there a clean way to solve an equation of this format algebraically? I can't seem to simplify the combination expression without using a large number of cases.
combinations factorial
combinations factorial
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
edited Jan 6 '15 at 18:12
Sujaan Kunalan
7,191133973
7,191133973
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
asked Jan 6 '15 at 18:06
user2520444user2520444
1816
1816
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Use the definition $$_nC_r = frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2cdot 3 cdot 7 cdot 11$ which means $$frac{12!}{924} =require{cancel} frac{1cdot 2cdot cancel{3} cdot cancel{4} cdot 5 cdot 6 cdot cancel{7} cdot 8 cdot 9 cdot 10 cdot cancel{11} cdot 12}{cancel{2^2}cdot cancel{3} cdot cancel{7} cdot cancel{11}} \ = 2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12$$ So now you have $$2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12=r!(12-r)!$$ Now you can quickly eliminate $r=0,1,2,3,4,5,7,8,9,10,11,12$ because if $r$ is one of those numbers, then either $r!$ or $(12-r)!$ will have a factor of $7$, and will not cancel with the LHS, hence you cannot have equality. This leaves you with $r=6$ as the only option.
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
$endgroup$
add a comment |
$begingroup$
Since $7$ is a factor of $frac{12!}{r!(12-r)!}=924$, $r!$ and $(12-r)!$ must not have factor $7$, which implies $r=6$.
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
$endgroup$
add a comment |
$begingroup$
I don't know of a quick way. As $12-r$ is a solution whenever $r$ is, you can concentrate on $r le frac {12}2$ For reasonably small values, you can then do binary search.
If the value $12$ is larger, you can use Stirling's approximation $12Cr=frac {12!}{r!(12-r)!}approxfrac {12^{12}sqrt{24pi}}{r^r(12-r)^{12-r}2pisqrt{r(12-r)}}$ Now take the logs and solve the equation to get $r=$ and expression involving $log r, log(12-r)$. Since the log changes rather slowly, guess that $r approx frac {12}4$ (the same starting point as the binary search. Plug that value in on the right, compute $r$, and you should come close with few iterations. Once you get close, use the fact that $r$ is a natural.
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
add a comment |
$begingroup$
You can spot the high prime factors of $924$ - which exist by Bertrand's Postulate.
Here you see $11$ and $7$ and the highest value for $r$ which avoids cancelling $7$ is $6$, which happens to be the only possibility.
This would work more generally to reduce the possbilities. Note that if a high prime doesn't appear in the factorisation that puts a constraint on too, because that prime has to cancel.
This puts the larger of $r$ and $n-r$ greater than or equal to the highest prime less than $n$ excluded from the factorisation and lower than any prime greater than $frac n2$ which is included.
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Use the definition $$_nC_r = frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2cdot 3 cdot 7 cdot 11$ which means $$frac{12!}{924} =require{cancel} frac{1cdot 2cdot cancel{3} cdot cancel{4} cdot 5 cdot 6 cdot cancel{7} cdot 8 cdot 9 cdot 10 cdot cancel{11} cdot 12}{cancel{2^2}cdot cancel{3} cdot cancel{7} cdot cancel{11}} \ = 2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12$$ So now you have $$2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12=r!(12-r)!$$ Now you can quickly eliminate $r=0,1,2,3,4,5,7,8,9,10,11,12$ because if $r$ is one of those numbers, then either $r!$ or $(12-r)!$ will have a factor of $7$, and will not cancel with the LHS, hence you cannot have equality. This leaves you with $r=6$ as the only option.
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
$endgroup$
add a comment |
$begingroup$
Use the definition $$_nC_r = frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2cdot 3 cdot 7 cdot 11$ which means $$frac{12!}{924} =require{cancel} frac{1cdot 2cdot cancel{3} cdot cancel{4} cdot 5 cdot 6 cdot cancel{7} cdot 8 cdot 9 cdot 10 cdot cancel{11} cdot 12}{cancel{2^2}cdot cancel{3} cdot cancel{7} cdot cancel{11}} \ = 2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12$$ So now you have $$2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12=r!(12-r)!$$ Now you can quickly eliminate $r=0,1,2,3,4,5,7,8,9,10,11,12$ because if $r$ is one of those numbers, then either $r!$ or $(12-r)!$ will have a factor of $7$, and will not cancel with the LHS, hence you cannot have equality. This leaves you with $r=6$ as the only option.
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
$endgroup$
add a comment |
$begingroup$
Use the definition $$_nC_r = frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2cdot 3 cdot 7 cdot 11$ which means $$frac{12!}{924} =require{cancel} frac{1cdot 2cdot cancel{3} cdot cancel{4} cdot 5 cdot 6 cdot cancel{7} cdot 8 cdot 9 cdot 10 cdot cancel{11} cdot 12}{cancel{2^2}cdot cancel{3} cdot cancel{7} cdot cancel{11}} \ = 2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12$$ So now you have $$2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12=r!(12-r)!$$ Now you can quickly eliminate $r=0,1,2,3,4,5,7,8,9,10,11,12$ because if $r$ is one of those numbers, then either $r!$ or $(12-r)!$ will have a factor of $7$, and will not cancel with the LHS, hence you cannot have equality. This leaves you with $r=6$ as the only option.
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
$endgroup$
Use the definition $$_nC_r = frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2cdot 3 cdot 7 cdot 11$ which means $$frac{12!}{924} =require{cancel} frac{1cdot 2cdot cancel{3} cdot cancel{4} cdot 5 cdot 6 cdot cancel{7} cdot 8 cdot 9 cdot 10 cdot cancel{11} cdot 12}{cancel{2^2}cdot cancel{3} cdot cancel{7} cdot cancel{11}} \ = 2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12$$ So now you have $$2 cdot 5 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12=r!(12-r)!$$ Now you can quickly eliminate $r=0,1,2,3,4,5,7,8,9,10,11,12$ because if $r$ is one of those numbers, then either $r!$ or $(12-r)!$ will have a factor of $7$, and will not cancel with the LHS, hence you cannot have equality. This leaves you with $r=6$ as the only option.
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
edited Jan 6 '15 at 19:20
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
answered Jan 6 '15 at 18:56
graydadgraydad
12.8k61933
12.8k61933
add a comment |
add a comment |
$begingroup$
Since $7$ is a factor of $frac{12!}{r!(12-r)!}=924$, $r!$ and $(12-r)!$ must not have factor $7$, which implies $r=6$.
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
$endgroup$
add a comment |
$begingroup$
Since $7$ is a factor of $frac{12!}{r!(12-r)!}=924$, $r!$ and $(12-r)!$ must not have factor $7$, which implies $r=6$.
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
$endgroup$
add a comment |
$begingroup$
Since $7$ is a factor of $frac{12!}{r!(12-r)!}=924$, $r!$ and $(12-r)!$ must not have factor $7$, which implies $r=6$.
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
$endgroup$
Since $7$ is a factor of $frac{12!}{r!(12-r)!}=924$, $r!$ and $(12-r)!$ must not have factor $7$, which implies $r=6$.
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
answered Jan 6 '15 at 18:26
zed111zed111
1,216721
1,216721
add a comment |
add a comment |
$begingroup$
I don't know of a quick way. As $12-r$ is a solution whenever $r$ is, you can concentrate on $r le frac {12}2$ For reasonably small values, you can then do binary search.
If the value $12$ is larger, you can use Stirling's approximation $12Cr=frac {12!}{r!(12-r)!}approxfrac {12^{12}sqrt{24pi}}{r^r(12-r)^{12-r}2pisqrt{r(12-r)}}$ Now take the logs and solve the equation to get $r=$ and expression involving $log r, log(12-r)$. Since the log changes rather slowly, guess that $r approx frac {12}4$ (the same starting point as the binary search. Plug that value in on the right, compute $r$, and you should come close with few iterations. Once you get close, use the fact that $r$ is a natural.
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
add a comment |
$begingroup$
I don't know of a quick way. As $12-r$ is a solution whenever $r$ is, you can concentrate on $r le frac {12}2$ For reasonably small values, you can then do binary search.
If the value $12$ is larger, you can use Stirling's approximation $12Cr=frac {12!}{r!(12-r)!}approxfrac {12^{12}sqrt{24pi}}{r^r(12-r)^{12-r}2pisqrt{r(12-r)}}$ Now take the logs and solve the equation to get $r=$ and expression involving $log r, log(12-r)$. Since the log changes rather slowly, guess that $r approx frac {12}4$ (the same starting point as the binary search. Plug that value in on the right, compute $r$, and you should come close with few iterations. Once you get close, use the fact that $r$ is a natural.
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
add a comment |
$begingroup$
I don't know of a quick way. As $12-r$ is a solution whenever $r$ is, you can concentrate on $r le frac {12}2$ For reasonably small values, you can then do binary search.
If the value $12$ is larger, you can use Stirling's approximation $12Cr=frac {12!}{r!(12-r)!}approxfrac {12^{12}sqrt{24pi}}{r^r(12-r)^{12-r}2pisqrt{r(12-r)}}$ Now take the logs and solve the equation to get $r=$ and expression involving $log r, log(12-r)$. Since the log changes rather slowly, guess that $r approx frac {12}4$ (the same starting point as the binary search. Plug that value in on the right, compute $r$, and you should come close with few iterations. Once you get close, use the fact that $r$ is a natural.
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
I don't know of a quick way. As $12-r$ is a solution whenever $r$ is, you can concentrate on $r le frac {12}2$ For reasonably small values, you can then do binary search.
If the value $12$ is larger, you can use Stirling's approximation $12Cr=frac {12!}{r!(12-r)!}approxfrac {12^{12}sqrt{24pi}}{r^r(12-r)^{12-r}2pisqrt{r(12-r)}}$ Now take the logs and solve the equation to get $r=$ and expression involving $log r, log(12-r)$. Since the log changes rather slowly, guess that $r approx frac {12}4$ (the same starting point as the binary search. Plug that value in on the right, compute $r$, and you should come close with few iterations. Once you get close, use the fact that $r$ is a natural.
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
answered Jan 6 '15 at 18:20
Ross MillikanRoss Millikan
298k24200374
298k24200374
add a comment |
add a comment |
$begingroup$
You can spot the high prime factors of $924$ - which exist by Bertrand's Postulate.
Here you see $11$ and $7$ and the highest value for $r$ which avoids cancelling $7$ is $6$, which happens to be the only possibility.
This would work more generally to reduce the possbilities. Note that if a high prime doesn't appear in the factorisation that puts a constraint on too, because that prime has to cancel.
This puts the larger of $r$ and $n-r$ greater than or equal to the highest prime less than $n$ excluded from the factorisation and lower than any prime greater than $frac n2$ which is included.
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
$endgroup$
add a comment |
$begingroup$
You can spot the high prime factors of $924$ - which exist by Bertrand's Postulate.
Here you see $11$ and $7$ and the highest value for $r$ which avoids cancelling $7$ is $6$, which happens to be the only possibility.
This would work more generally to reduce the possbilities. Note that if a high prime doesn't appear in the factorisation that puts a constraint on too, because that prime has to cancel.
This puts the larger of $r$ and $n-r$ greater than or equal to the highest prime less than $n$ excluded from the factorisation and lower than any prime greater than $frac n2$ which is included.
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
$endgroup$
add a comment |
$begingroup$
You can spot the high prime factors of $924$ - which exist by Bertrand's Postulate.
Here you see $11$ and $7$ and the highest value for $r$ which avoids cancelling $7$ is $6$, which happens to be the only possibility.
This would work more generally to reduce the possbilities. Note that if a high prime doesn't appear in the factorisation that puts a constraint on too, because that prime has to cancel.
This puts the larger of $r$ and $n-r$ greater than or equal to the highest prime less than $n$ excluded from the factorisation and lower than any prime greater than $frac n2$ which is included.
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
$endgroup$
You can spot the high prime factors of $924$ - which exist by Bertrand's Postulate.
Here you see $11$ and $7$ and the highest value for $r$ which avoids cancelling $7$ is $6$, which happens to be the only possibility.
This would work more generally to reduce the possbilities. Note that if a high prime doesn't appear in the factorisation that puts a constraint on too, because that prime has to cancel.
This puts the larger of $r$ and $n-r$ greater than or equal to the highest prime less than $n$ excluded from the factorisation and lower than any prime greater than $frac n2$ which is included.
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
edited Jan 6 '15 at 20:09
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
answered Jan 6 '15 at 18:25
Mark BennetMark Bennet
81.5k983180
81.5k983180
add a comment |
add a comment |
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