Mixing
Solving for r in a combinatorial problem
Clash Royale CLAN TAG#URR8PPP
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Find the value of $r$:
$$frac1dbinom9r - frac1dbinom10r = frac116dbinom11r$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many factorials that don't seem to combine.
combinatorics
edited 5 hours ago
N. F. Taussig
40k93253
asked 5 hours ago
Nick R
161
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Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote
favorite
Find the value of $r$:
$$frac1dbinom9r - frac1dbinom10r = frac116dbinom11r$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many factorials that don't seem to combine.
combinatorics
edited 5 hours ago
N. F. Taussig
40k93253
asked 5 hours ago
Nick R
161
New contributor
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. When you pose a question here, you should show what you have attempted in addition to explaining where you got stuck. That helps users identify any mistakes you may have made and write answers appropriate to your skill level.
– N. F. Taussig
5 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find the value of $r$:
$$frac1dbinom9r - frac1dbinom10r = frac116dbinom11r$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many factorials that don't seem to combine.
combinatorics
edited 5 hours ago
N. F. Taussig
40k93253
asked 5 hours ago
Nick R
161
New contributor
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Find the value of $r$:
$$frac1dbinom9r - frac1dbinom10r = frac116dbinom11r$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many factorials that don't seem to combine.
combinatorics
combinatorics
edited 5 hours ago
N. F. Taussig
40k93253
asked 5 hours ago
Nick R
161
New contributor
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
N. F. Taussig
40k93253
asked 5 hours ago
Nick R
161
New contributor
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
N. F. Taussig
40k93253
edited 5 hours ago
N. F. Taussig
40k93253
edited 5 hours ago
N. F. Taussig
40k93253
40k93253
asked 5 hours ago
Nick R
161
New contributor
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Nick R
161
asked 5 hours ago
Nick R
161
161
New contributor
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nick R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. When you pose a question here, you should show what you have attempted in addition to explaining where you got stuck. That helps users identify any mistakes you may have made and write answers appropriate to your skill level.
– N. F. Taussig
5 hours ago
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Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. When you pose a question here, you should show what you have attempted in addition to explaining where you got stuck. That helps users identify any mistakes you may have made and write answers appropriate to your skill level.
– N. F. Taussig
5 hours ago
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. When you pose a question here, you should show what you have attempted in addition to explaining where you got stuck. That helps users identify any mistakes you may have made and write answers appropriate to your skill level.
– N. F. Taussig
5 hours ago
Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. When you pose a question here, you should show what you have attempted in addition to explaining where you got stuck. That helps users identify any mistakes you may have made and write answers appropriate to your skill level.
– N. F. Taussig
5 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Since
$$binomnk =
begincases
dfracn!k!(n - k)! & textif $0 leq k leq n$\[2 mm]
0 & textif $k > n$
endcases$$
we require that $k leq 9$ since otherwise we would be dividing by $0$.
beginalign*
frac1binom9k - frac1binom10k & = frac116binom11k\
frac1dfrac9!k!(9 - k)! - frac1dfrac10!k!(10 - k)! & = frac116 cdot dfrac11!k!(11 - k)! && textby definition of $binomnk$\
frack!(9 - k)!9! - frack!(10 - k)!10! & = frac11k!(11 - k)!6 cdot 11! && textdivision is multiplication by the reciprocal\
frac10k!(9 - k)!10! - frack!(10 - k)!10! & = frack!(11 - k)!6 cdot 10! && textform common denominator on LHS, cancel on RHS\
6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && textmultiply by $6 cdot 10!$\
6[10 - (10 - k)] & = (11 - k)(10 - k) && textdivide by $k!(9 - k)!$\
6k & = 110 - 21k + k^2 && textsimplify\
0 & = k^2 - 27k + 110 && textset quadratic equal to zero
endalign*
When you solve the quadratic equation, keep in mind the restriction that $k leq 9$.
answered 4 hours ago
N. F. Taussig
40k93253
add a comment |Â
up vote
3
down vote
Hopefully you know the formula $dbinomnk = fracn!(n-k)!k!$.
So your equation becomes:
$$fracr!(9-r)!9! - fracr!(10-r)!10! = frac11times r!(11-r)!6times 11!$$
$$fracr!(9-r)!9! - fracr!(9-r)!(10-r)9!times 10 = frac11times r!(9-r)!(10-r)(11-r)6times 9!times10times11$$
Divide both sides by $fracr!(9-r)!9!$:
$$1-frac10-r10 = frac11times (10-r)(11-r)6times 10times11$$
I hope it's now clear that this is a quadratic equation. I haven't tried to solve it, but the work should be simple.
EDIT: I totally forgot about the restriction. Taussig's answer is correct, after solving the quadratic equation you must eliminate any r>9.
answered 5 hours ago
Felix Fourcolor
787
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Since
$$binomnk =
begincases
dfracn!k!(n - k)! & textif $0 leq k leq n$\[2 mm]
0 & textif $k > n$
endcases$$
we require that $k leq 9$ since otherwise we would be dividing by $0$.
beginalign*
frac1binom9k - frac1binom10k & = frac116binom11k\
frac1dfrac9!k!(9 - k)! - frac1dfrac10!k!(10 - k)! & = frac116 cdot dfrac11!k!(11 - k)! && textby definition of $binomnk$\
frack!(9 - k)!9! - frack!(10 - k)!10! & = frac11k!(11 - k)!6 cdot 11! && textdivision is multiplication by the reciprocal\
frac10k!(9 - k)!10! - frack!(10 - k)!10! & = frack!(11 - k)!6 cdot 10! && textform common denominator on LHS, cancel on RHS\
6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && textmultiply by $6 cdot 10!$\
6[10 - (10 - k)] & = (11 - k)(10 - k) && textdivide by $k!(9 - k)!$\
6k & = 110 - 21k + k^2 && textsimplify\
0 & = k^2 - 27k + 110 && textset quadratic equal to zero
endalign*
When you solve the quadratic equation, keep in mind the restriction that $k leq 9$.
answered 4 hours ago
N. F. Taussig
40k93253
add a comment |Â
up vote
3
down vote
Since
$$binomnk =
begincases
dfracn!k!(n - k)! & textif $0 leq k leq n$\[2 mm]
0 & textif $k > n$
endcases$$
we require that $k leq 9$ since otherwise we would be dividing by $0$.
beginalign*
frac1binom9k - frac1binom10k & = frac116binom11k\
frac1dfrac9!k!(9 - k)! - frac1dfrac10!k!(10 - k)! & = frac116 cdot dfrac11!k!(11 - k)! && textby definition of $binomnk$\
frack!(9 - k)!9! - frack!(10 - k)!10! & = frac11k!(11 - k)!6 cdot 11! && textdivision is multiplication by the reciprocal\
frac10k!(9 - k)!10! - frack!(10 - k)!10! & = frack!(11 - k)!6 cdot 10! && textform common denominator on LHS, cancel on RHS\
6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && textmultiply by $6 cdot 10!$\
6[10 - (10 - k)] & = (11 - k)(10 - k) && textdivide by $k!(9 - k)!$\
6k & = 110 - 21k + k^2 && textsimplify\
0 & = k^2 - 27k + 110 && textset quadratic equal to zero
endalign*
When you solve the quadratic equation, keep in mind the restriction that $k leq 9$.
answered 4 hours ago
N. F. Taussig
40k93253
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Since
$$binomnk =
begincases
dfracn!k!(n - k)! & textif $0 leq k leq n$\[2 mm]
0 & textif $k > n$
endcases$$
we require that $k leq 9$ since otherwise we would be dividing by $0$.
beginalign*
frac1binom9k - frac1binom10k & = frac116binom11k\
frac1dfrac9!k!(9 - k)! - frac1dfrac10!k!(10 - k)! & = frac116 cdot dfrac11!k!(11 - k)! && textby definition of $binomnk$\
frack!(9 - k)!9! - frack!(10 - k)!10! & = frac11k!(11 - k)!6 cdot 11! && textdivision is multiplication by the reciprocal\
frac10k!(9 - k)!10! - frack!(10 - k)!10! & = frack!(11 - k)!6 cdot 10! && textform common denominator on LHS, cancel on RHS\
6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && textmultiply by $6 cdot 10!$\
6[10 - (10 - k)] & = (11 - k)(10 - k) && textdivide by $k!(9 - k)!$\
6k & = 110 - 21k + k^2 && textsimplify\
0 & = k^2 - 27k + 110 && textset quadratic equal to zero
endalign*
When you solve the quadratic equation, keep in mind the restriction that $k leq 9$.
answered 4 hours ago
N. F. Taussig
40k93253
Since
$$binomnk =
begincases
dfracn!k!(n - k)! & textif $0 leq k leq n$\[2 mm]
0 & textif $k > n$
endcases$$
we require that $k leq 9$ since otherwise we would be dividing by $0$.
beginalign*
frac1binom9k - frac1binom10k & = frac116binom11k\
frac1dfrac9!k!(9 - k)! - frac1dfrac10!k!(10 - k)! & = frac116 cdot dfrac11!k!(11 - k)! && textby definition of $binomnk$\
frack!(9 - k)!9! - frack!(10 - k)!10! & = frac11k!(11 - k)!6 cdot 11! && textdivision is multiplication by the reciprocal\
frac10k!(9 - k)!10! - frack!(10 - k)!10! & = frack!(11 - k)!6 cdot 10! && textform common denominator on LHS, cancel on RHS\
6[10k!(9 - k)! - k!(10 - k)!] & = k!(11 - k)! && textmultiply by $6 cdot 10!$\
6[10 - (10 - k)] & = (11 - k)(10 - k) && textdivide by $k!(9 - k)!$\
6k & = 110 - 21k + k^2 && textsimplify\
0 & = k^2 - 27k + 110 && textset quadratic equal to zero
endalign*
When you solve the quadratic equation, keep in mind the restriction that $k leq 9$.
answered 4 hours ago
N. F. Taussig
40k93253
answered 4 hours ago
N. F. Taussig
40k93253
answered 4 hours ago
N. F. Taussig
40k93253
answered 4 hours ago
N. F. Taussig
40k93253
40k93253
add a comment |Â
add a comment |Â
up vote
3
down vote
Hopefully you know the formula $dbinomnk = fracn!(n-k)!k!$.
So your equation becomes:
$$fracr!(9-r)!9! - fracr!(10-r)!10! = frac11times r!(11-r)!6times 11!$$
$$fracr!(9-r)!9! - fracr!(9-r)!(10-r)9!times 10 = frac11times r!(9-r)!(10-r)(11-r)6times 9!times10times11$$
Divide both sides by $fracr!(9-r)!9!$:
$$1-frac10-r10 = frac11times (10-r)(11-r)6times 10times11$$
I hope it's now clear that this is a quadratic equation. I haven't tried to solve it, but the work should be simple.
EDIT: I totally forgot about the restriction. Taussig's answer is correct, after solving the quadratic equation you must eliminate any r>9.
answered 5 hours ago
Felix Fourcolor
787
add a comment |Â
up vote
3
down vote
Hopefully you know the formula $dbinomnk = fracn!(n-k)!k!$.
So your equation becomes:
$$fracr!(9-r)!9! - fracr!(10-r)!10! = frac11times r!(11-r)!6times 11!$$
$$fracr!(9-r)!9! - fracr!(9-r)!(10-r)9!times 10 = frac11times r!(9-r)!(10-r)(11-r)6times 9!times10times11$$
Divide both sides by $fracr!(9-r)!9!$:
$$1-frac10-r10 = frac11times (10-r)(11-r)6times 10times11$$
I hope it's now clear that this is a quadratic equation. I haven't tried to solve it, but the work should be simple.
EDIT: I totally forgot about the restriction. Taussig's answer is correct, after solving the quadratic equation you must eliminate any r>9.
answered 5 hours ago
Felix Fourcolor
787
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hopefully you know the formula $dbinomnk = fracn!(n-k)!k!$.
So your equation becomes:
$$fracr!(9-r)!9! - fracr!(10-r)!10! = frac11times r!(11-r)!6times 11!$$
$$fracr!(9-r)!9! - fracr!(9-r)!(10-r)9!times 10 = frac11times r!(9-r)!(10-r)(11-r)6times 9!times10times11$$
Divide both sides by $fracr!(9-r)!9!$:
$$1-frac10-r10 = frac11times (10-r)(11-r)6times 10times11$$
I hope it's now clear that this is a quadratic equation. I haven't tried to solve it, but the work should be simple.
EDIT: I totally forgot about the restriction. Taussig's answer is correct, after solving the quadratic equation you must eliminate any r>9.
answered 5 hours ago
Felix Fourcolor
787
Hopefully you know the formula $dbinomnk = fracn!(n-k)!k!$.
So your equation becomes:
$$fracr!(9-r)!9! - fracr!(10-r)!10! = frac11times r!(11-r)!6times 11!$$
$$fracr!(9-r)!9! - fracr!(9-r)!(10-r)9!times 10 = frac11times r!(9-r)!(10-r)(11-r)6times 9!times10times11$$
Divide both sides by $fracr!(9-r)!9!$:
$$1-frac10-r10 = frac11times (10-r)(11-r)6times 10times11$$
I hope it's now clear that this is a quadratic equation. I haven't tried to solve it, but the work should be simple.
EDIT: I totally forgot about the restriction. Taussig's answer is correct, after solving the quadratic equation you must eliminate any r>9.
answered 5 hours ago
Felix Fourcolor
787
edited 1 hour ago
answered 5 hours ago
Felix Fourcolor
787
answered 5 hours ago
Felix Fourcolor
787
answered 5 hours ago
Felix Fourcolor
787
787
add a comment |Â
add a comment |Â
Nick R is a new contributor. Be nice, and check out our Code of Conduct.
Nick R is a new contributor. Be nice, and check out our Code of Conduct.
Nick R is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. When you pose a question here, you should show what you have attempted in addition to explaining where you got stuck. That helps users identify any mistakes you may have made and write answers appropriate to your skill level.
– N. F. Taussig
5 hours ago