Mixing
How to give k identical objects to n different people such that each of them gets odd number of these...
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How to give k identical objects to n different people such that each of them gets odd number of these objects? I don’t know how to include odd number thing.
combinatorics
asked Jan 23 at 18:47
MichaelMichael
246
$endgroup$
add a comment |
$begingroup$
How to give k identical objects to n different people such that each of them gets odd number of these objects? I don’t know how to include odd number thing.
combinatorics
asked Jan 23 at 18:47
MichaelMichael
246
$endgroup$
$begingroup$
Show your work: can you find the number of way to distribute the objects without the restriction?
$endgroup$
– Daniel Mathias
Jan 23 at 18:49
$begingroup$
HINT Given each person $1$ ... and hand out the rest in pairs. How many such pairs are there?
$endgroup$
– Bram28
Jan 23 at 18:50
$begingroup$
If n is number of peopl and k number of objects then (k+n-1) choose (n-1)
$endgroup$
– Michael
Jan 23 at 18:51
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You may be asking about "partitions into odd parts," but I'm not sure.
$endgroup$
– saulspatz
Jan 23 at 18:51
add a comment |
$begingroup$
How to give k identical objects to n different people such that each of them gets odd number of these objects? I don’t know how to include odd number thing.
combinatorics
asked Jan 23 at 18:47
MichaelMichael
246
$endgroup$
How to give k identical objects to n different people such that each of them gets odd number of these objects? I don’t know how to include odd number thing.
combinatorics
combinatorics
asked Jan 23 at 18:47
MichaelMichael
246
asked Jan 23 at 18:47
MichaelMichael
246
asked Jan 23 at 18:47
MichaelMichael
246
asked Jan 23 at 18:47
MichaelMichael
246
asked Jan 23 at 18:47
MichaelMichael
246
246
$begingroup$
Show your work: can you find the number of way to distribute the objects without the restriction?
$endgroup$
– Daniel Mathias
Jan 23 at 18:49
$begingroup$
HINT Given each person $1$ ... and hand out the rest in pairs. How many such pairs are there?
$endgroup$
– Bram28
Jan 23 at 18:50
$begingroup$
If n is number of peopl and k number of objects then (k+n-1) choose (n-1)
$endgroup$
– Michael
Jan 23 at 18:51
$begingroup$
You may be asking about "partitions into odd parts," but I'm not sure.
$endgroup$
– saulspatz
Jan 23 at 18:51
add a comment |
$begingroup$
Show your work: can you find the number of way to distribute the objects without the restriction?
$endgroup$
– Daniel Mathias
Jan 23 at 18:49
$begingroup$
HINT Given each person $1$ ... and hand out the rest in pairs. How many such pairs are there?
$endgroup$
– Bram28
Jan 23 at 18:50
$begingroup$
If n is number of peopl and k number of objects then (k+n-1) choose (n-1)
$endgroup$
– Michael
Jan 23 at 18:51
$begingroup$
You may be asking about "partitions into odd parts," but I'm not sure.
$endgroup$
– saulspatz
Jan 23 at 18:51
$begingroup$
Show your work: can you find the number of way to distribute the objects without the restriction?
$endgroup$
– Daniel Mathias
Jan 23 at 18:49
$begingroup$
Show your work: can you find the number of way to distribute the objects without the restriction?
$endgroup$
– Daniel Mathias
Jan 23 at 18:49
$begingroup$
HINT Given each person $1$ ... and hand out the rest in pairs. How many such pairs are there?
$endgroup$
– Bram28
Jan 23 at 18:50
$begingroup$
HINT Given each person $1$ ... and hand out the rest in pairs. How many such pairs are there?
$endgroup$
– Bram28
Jan 23 at 18:50
$begingroup$
If n is number of peopl and k number of objects then (k+n-1) choose (n-1)
$endgroup$
– Michael
Jan 23 at 18:51
$begingroup$
If n is number of peopl and k number of objects then (k+n-1) choose (n-1)
$endgroup$
– Michael
Jan 23 at 18:51
$begingroup$
You may be asking about "partitions into odd parts," but I'm not sure.
$endgroup$
– saulspatz
Jan 23 at 18:51
$begingroup$
You may be asking about "partitions into odd parts," but I'm not sure.
$endgroup$
– saulspatz
Jan 23 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Letting $x_i$ be the number of objects given to person $i$, you are counting soluitions to the equation
$$
x_1+x_2+dots+x_n = k,
$$
subject to the constraint that each $x_i$ is an odd positive integer. Letting $y_i=(x_i-1)/2$, this is equivalent to counting nonnegative integer solutions to
$$
y_1+y_2+dots+y_n=(k-n)/2,
$$
which can be done with stars and bars.
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
$endgroup$
$begingroup$
How can I use stars and bars then?
$endgroup$
– Michael
Jan 23 at 19:14
1
$begingroup$
@Michael you already gave the solution in your response to my comment: $binom{k+n-1}{n-1}$ becomes $binom{(k-n)/2+n-1}{n-1}$
$endgroup$
– Daniel Mathias
Jan 23 at 19:16
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@DanielMathias Thanks for pointing out the error!
$endgroup$
– Mike Earnest
Jan 23 at 19:17
$begingroup$
Thank you for your solution! Now I got it.
$endgroup$
– Michael
Jan 23 at 19:20
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
$begingroup$
Letting $x_i$ be the number of objects given to person $i$, you are counting soluitions to the equation
$$
x_1+x_2+dots+x_n = k,
$$
subject to the constraint that each $x_i$ is an odd positive integer. Letting $y_i=(x_i-1)/2$, this is equivalent to counting nonnegative integer solutions to
$$
y_1+y_2+dots+y_n=(k-n)/2,
$$
which can be done with stars and bars.
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
$endgroup$
$begingroup$
How can I use stars and bars then?
$endgroup$
– Michael
Jan 23 at 19:14
1
$begingroup$
@Michael you already gave the solution in your response to my comment: $binom{k+n-1}{n-1}$ becomes $binom{(k-n)/2+n-1}{n-1}$
$endgroup$
– Daniel Mathias
Jan 23 at 19:16
$begingroup$
@DanielMathias Thanks for pointing out the error!
$endgroup$
– Mike Earnest
Jan 23 at 19:17
$begingroup$
Thank you for your solution! Now I got it.
$endgroup$
– Michael
Jan 23 at 19:20
add a comment |
$begingroup$
Letting $x_i$ be the number of objects given to person $i$, you are counting soluitions to the equation
$$
x_1+x_2+dots+x_n = k,
$$
subject to the constraint that each $x_i$ is an odd positive integer. Letting $y_i=(x_i-1)/2$, this is equivalent to counting nonnegative integer solutions to
$$
y_1+y_2+dots+y_n=(k-n)/2,
$$
which can be done with stars and bars.
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
$endgroup$
$begingroup$
How can I use stars and bars then?
$endgroup$
– Michael
Jan 23 at 19:14
1
$begingroup$
@Michael you already gave the solution in your response to my comment: $binom{k+n-1}{n-1}$ becomes $binom{(k-n)/2+n-1}{n-1}$
$endgroup$
– Daniel Mathias
Jan 23 at 19:16
$begingroup$
@DanielMathias Thanks for pointing out the error!
$endgroup$
– Mike Earnest
Jan 23 at 19:17
$begingroup$
Thank you for your solution! Now I got it.
$endgroup$
– Michael
Jan 23 at 19:20
add a comment |
$begingroup$
Letting $x_i$ be the number of objects given to person $i$, you are counting soluitions to the equation
$$
x_1+x_2+dots+x_n = k,
$$
subject to the constraint that each $x_i$ is an odd positive integer. Letting $y_i=(x_i-1)/2$, this is equivalent to counting nonnegative integer solutions to
$$
y_1+y_2+dots+y_n=(k-n)/2,
$$
which can be done with stars and bars.
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
$endgroup$
Letting $x_i$ be the number of objects given to person $i$, you are counting soluitions to the equation
$$
x_1+x_2+dots+x_n = k,
$$
subject to the constraint that each $x_i$ is an odd positive integer. Letting $y_i=(x_i-1)/2$, this is equivalent to counting nonnegative integer solutions to
$$
y_1+y_2+dots+y_n=(k-n)/2,
$$
which can be done with stars and bars.
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
edited Jan 23 at 19:14
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
answered Jan 23 at 19:10
Mike EarnestMike Earnest
24.6k22151
24.6k22151
$begingroup$
How can I use stars and bars then?
$endgroup$
– Michael
Jan 23 at 19:14
1
$begingroup$
@Michael you already gave the solution in your response to my comment: $binom{k+n-1}{n-1}$ becomes $binom{(k-n)/2+n-1}{n-1}$
$endgroup$
– Daniel Mathias
Jan 23 at 19:16
$begingroup$
@DanielMathias Thanks for pointing out the error!
$endgroup$
– Mike Earnest
Jan 23 at 19:17
$begingroup$
Thank you for your solution! Now I got it.
$endgroup$
– Michael
Jan 23 at 19:20
add a comment |
$begingroup$
How can I use stars and bars then?
$endgroup$
– Michael
Jan 23 at 19:14
1
$begingroup$
@Michael you already gave the solution in your response to my comment: $binom{k+n-1}{n-1}$ becomes $binom{(k-n)/2+n-1}{n-1}$
$endgroup$
– Daniel Mathias
Jan 23 at 19:16
$begingroup$
@DanielMathias Thanks for pointing out the error!
$endgroup$
– Mike Earnest
Jan 23 at 19:17
$begingroup$
Thank you for your solution! Now I got it.
$endgroup$
– Michael
Jan 23 at 19:20
$begingroup$
How can I use stars and bars then?
$endgroup$
– Michael
Jan 23 at 19:14
$begingroup$
How can I use stars and bars then?
$endgroup$
– Michael
Jan 23 at 19:14
1
1
$begingroup$
@Michael you already gave the solution in your response to my comment: $binom{k+n-1}{n-1}$ becomes $binom{(k-n)/2+n-1}{n-1}$
$endgroup$
– Daniel Mathias
Jan 23 at 19:16
$begingroup$
@Michael you already gave the solution in your response to my comment: $binom{k+n-1}{n-1}$ becomes $binom{(k-n)/2+n-1}{n-1}$
$endgroup$
– Daniel Mathias
Jan 23 at 19:16
$begingroup$
@DanielMathias Thanks for pointing out the error!
$endgroup$
– Mike Earnest
Jan 23 at 19:17
$begingroup$
@DanielMathias Thanks for pointing out the error!
$endgroup$
– Mike Earnest
Jan 23 at 19:17
$begingroup$
Thank you for your solution! Now I got it.
$endgroup$
– Michael
Jan 23 at 19:20
$begingroup$
Thank you for your solution! Now I got it.
$endgroup$
– Michael
Jan 23 at 19:20
add a comment |
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$begingroup$
Show your work: can you find the number of way to distribute the objects without the restriction?
$endgroup$
– Daniel Mathias
Jan 23 at 18:49
$begingroup$
HINT Given each person $1$ ... and hand out the rest in pairs. How many such pairs are there?
$endgroup$
– Bram28
Jan 23 at 18:50
$begingroup$
If n is number of peopl and k number of objects then (k+n-1) choose (n-1)
$endgroup$
– Michael
Jan 23 at 18:51
$begingroup$
You may be asking about "partitions into odd parts," but I'm not sure.
$endgroup$
– saulspatz
Jan 23 at 18:51