Mixing
I know numbers of four items and want to find out possible number of pairs
$begingroup$
I have 4 kinds of items a,b,c,d, and I know numbers of these four items: n1,n2,n3,n4. These items form pairs with each other and items of the same kind can also form a pair:ab, aa, ac, cd...Order doesn't matter, so ab and ba are the same. Now I don't know how many pairs are there between these items. Is there any possible way to calculate possible number of each pairs?
probability statistics
asked Jan 22 at 18:32
AmandaAmanda
62
$endgroup$
add a comment |
$begingroup$
I have 4 kinds of items a,b,c,d, and I know numbers of these four items: n1,n2,n3,n4. These items form pairs with each other and items of the same kind can also form a pair:ab, aa, ac, cd...Order doesn't matter, so ab and ba are the same. Now I don't know how many pairs are there between these items. Is there any possible way to calculate possible number of each pairs?
probability statistics
asked Jan 22 at 18:32
AmandaAmanda
62
$endgroup$
1
$begingroup$
The fact that the items are indistinct makes this question a little vague. You could make the argument that the answer is 10 regardless of what $n_1,ldots,n_4$ are: $aa,bb,cc,dd,ab,ac,ad,bc,bd,cd$. Unless you are asking in how many ways you can pair up ALL of the elements?
$endgroup$
– Nick Peterson
Jan 22 at 19:09
$begingroup$
would like to get your feedback on the answer I provided.
$endgroup$
– G Cab
Jan 28 at 22:26
add a comment |
$begingroup$
I have 4 kinds of items a,b,c,d, and I know numbers of these four items: n1,n2,n3,n4. These items form pairs with each other and items of the same kind can also form a pair:ab, aa, ac, cd...Order doesn't matter, so ab and ba are the same. Now I don't know how many pairs are there between these items. Is there any possible way to calculate possible number of each pairs?
probability statistics
asked Jan 22 at 18:32
AmandaAmanda
62
$endgroup$
I have 4 kinds of items a,b,c,d, and I know numbers of these four items: n1,n2,n3,n4. These items form pairs with each other and items of the same kind can also form a pair:ab, aa, ac, cd...Order doesn't matter, so ab and ba are the same. Now I don't know how many pairs are there between these items. Is there any possible way to calculate possible number of each pairs?
probability statistics
probability statistics
asked Jan 22 at 18:32
AmandaAmanda
62
asked Jan 22 at 18:32
AmandaAmanda
62
asked Jan 22 at 18:32
AmandaAmanda
62
asked Jan 22 at 18:32
AmandaAmanda
62
asked Jan 22 at 18:32
AmandaAmanda
62
62
1
$begingroup$
The fact that the items are indistinct makes this question a little vague. You could make the argument that the answer is 10 regardless of what $n_1,ldots,n_4$ are: $aa,bb,cc,dd,ab,ac,ad,bc,bd,cd$. Unless you are asking in how many ways you can pair up ALL of the elements?
$endgroup$
– Nick Peterson
Jan 22 at 19:09
$begingroup$
would like to get your feedback on the answer I provided.
$endgroup$
– G Cab
Jan 28 at 22:26
add a comment |
1
$begingroup$
The fact that the items are indistinct makes this question a little vague. You could make the argument that the answer is 10 regardless of what $n_1,ldots,n_4$ are: $aa,bb,cc,dd,ab,ac,ad,bc,bd,cd$. Unless you are asking in how many ways you can pair up ALL of the elements?
$endgroup$
– Nick Peterson
Jan 22 at 19:09
$begingroup$
would like to get your feedback on the answer I provided.
$endgroup$
– G Cab
Jan 28 at 22:26
1
1
$begingroup$
The fact that the items are indistinct makes this question a little vague. You could make the argument that the answer is 10 regardless of what $n_1,ldots,n_4$ are: $aa,bb,cc,dd,ab,ac,ad,bc,bd,cd$. Unless you are asking in how many ways you can pair up ALL of the elements?
$endgroup$
– Nick Peterson
Jan 22 at 19:09
$begingroup$
The fact that the items are indistinct makes this question a little vague. You could make the argument that the answer is 10 regardless of what $n_1,ldots,n_4$ are: $aa,bb,cc,dd,ab,ac,ad,bc,bd,cd$. Unless you are asking in how many ways you can pair up ALL of the elements?
$endgroup$
– Nick Peterson
Jan 22 at 19:09
$begingroup$
would like to get your feedback on the answer I provided.
$endgroup$
– G Cab
Jan 28 at 22:26
$begingroup$
would like to get your feedback on the answer I provided.
$endgroup$
– G Cab
Jan 28 at 22:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have $n_1$ items of type $a$, $n_2$ of type $b$, $n_3$ of type $c$, and $n_4$ of type $d$. Individual items of a given type are indistinct. You can choose $2$ items of one type or $1$ each of two types. The number of ways to choose two items of specified type is:
$$aa:binom{n_1}{2}quad bb:binom{n_2}{2}quad cc:binom{n_3}{2}quad dd:binom{n_4}{2}\[4ex]ab:n_1times n_2quad ac:n_1times n_3quad ad:n_1times n_4\[2ex]bc:n_2times n_3quad bd:n_2times n_4quad cd:n_3times n_4$$
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
$endgroup$
add a comment |
$begingroup$
Let me introduce what is just a first step.
We shall first of all assume that
$$
n_{;1} + n_{;2} + n_{;3} + n_{;4} = 2m
$$
Then, since order doesn't matter, we are going to arrange the couples alphabetically
$$
left[ {underbrace {aa, ldots ,aa}_{c_{,1,,1} },underbrace {ab, ldots ,ab}_{c_{,1,,2} },
cdots ,underbrace {bb, ldots }_{c_{,2,,2} }, cdots ; cdots ,cd,underbrace {dd, ldots ,dd}_{c_{,4,,4} }} right]
$$
and indicate the number of each different couple by $c_{k,j}$ as shown.
Therefore the symmetric matrix made up by them as shown, shall satisfy
$$
left( {matrix{
{2c_{,1,,1} } & {c_{,1,,2} } & {c_{,1,,3} } & {c_{,1,,4} } cr
{c_{,2,,1} } & {2c_{,2,,2} } & {c_{,2,,3} } & {c_{,2,,4} } cr
{c_{,3,,1} } & {c_{,3,,2} } & {2c_{,3,,3} } & {c_{,3,,4} } cr
{c_{,4,,1} } & {c_{,4,,2} } & {c_{,4,,3} } & {2c_{,4,,4} } cr
} } right)left( {matrix{ 1 cr 1 cr 1 cr 1 cr
} } right) = left( {matrix{ {n_{;1} } cr {n_{;2} } cr {n_{;3} } cr {n_{;4} } cr
} } right)quad left| {;c_{,j,,k} = c_{,k,,j} } right.
$$
which is a diophantine linear system of four equations in ten unknowns,
which are required to be non-negative integers.
We are interested in the number of solutions of such a system.
It is not an easy task, as you can see from the refered link.
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
You have $n_1$ items of type $a$, $n_2$ of type $b$, $n_3$ of type $c$, and $n_4$ of type $d$. Individual items of a given type are indistinct. You can choose $2$ items of one type or $1$ each of two types. The number of ways to choose two items of specified type is:
$$aa:binom{n_1}{2}quad bb:binom{n_2}{2}quad cc:binom{n_3}{2}quad dd:binom{n_4}{2}\[4ex]ab:n_1times n_2quad ac:n_1times n_3quad ad:n_1times n_4\[2ex]bc:n_2times n_3quad bd:n_2times n_4quad cd:n_3times n_4$$
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
$endgroup$
add a comment |
$begingroup$
You have $n_1$ items of type $a$, $n_2$ of type $b$, $n_3$ of type $c$, and $n_4$ of type $d$. Individual items of a given type are indistinct. You can choose $2$ items of one type or $1$ each of two types. The number of ways to choose two items of specified type is:
$$aa:binom{n_1}{2}quad bb:binom{n_2}{2}quad cc:binom{n_3}{2}quad dd:binom{n_4}{2}\[4ex]ab:n_1times n_2quad ac:n_1times n_3quad ad:n_1times n_4\[2ex]bc:n_2times n_3quad bd:n_2times n_4quad cd:n_3times n_4$$
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
$endgroup$
add a comment |
$begingroup$
You have $n_1$ items of type $a$, $n_2$ of type $b$, $n_3$ of type $c$, and $n_4$ of type $d$. Individual items of a given type are indistinct. You can choose $2$ items of one type or $1$ each of two types. The number of ways to choose two items of specified type is:
$$aa:binom{n_1}{2}quad bb:binom{n_2}{2}quad cc:binom{n_3}{2}quad dd:binom{n_4}{2}\[4ex]ab:n_1times n_2quad ac:n_1times n_3quad ad:n_1times n_4\[2ex]bc:n_2times n_3quad bd:n_2times n_4quad cd:n_3times n_4$$
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
$endgroup$
You have $n_1$ items of type $a$, $n_2$ of type $b$, $n_3$ of type $c$, and $n_4$ of type $d$. Individual items of a given type are indistinct. You can choose $2$ items of one type or $1$ each of two types. The number of ways to choose two items of specified type is:
$$aa:binom{n_1}{2}quad bb:binom{n_2}{2}quad cc:binom{n_3}{2}quad dd:binom{n_4}{2}\[4ex]ab:n_1times n_2quad ac:n_1times n_3quad ad:n_1times n_4\[2ex]bc:n_2times n_3quad bd:n_2times n_4quad cd:n_3times n_4$$
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
answered Jan 22 at 19:03
Daniel MathiasDaniel Mathias
1,35018
1,35018
add a comment |
add a comment |
$begingroup$
Let me introduce what is just a first step.
We shall first of all assume that
$$
n_{;1} + n_{;2} + n_{;3} + n_{;4} = 2m
$$
Then, since order doesn't matter, we are going to arrange the couples alphabetically
$$
left[ {underbrace {aa, ldots ,aa}_{c_{,1,,1} },underbrace {ab, ldots ,ab}_{c_{,1,,2} },
cdots ,underbrace {bb, ldots }_{c_{,2,,2} }, cdots ; cdots ,cd,underbrace {dd, ldots ,dd}_{c_{,4,,4} }} right]
$$
and indicate the number of each different couple by $c_{k,j}$ as shown.
Therefore the symmetric matrix made up by them as shown, shall satisfy
$$
left( {matrix{
{2c_{,1,,1} } & {c_{,1,,2} } & {c_{,1,,3} } & {c_{,1,,4} } cr
{c_{,2,,1} } & {2c_{,2,,2} } & {c_{,2,,3} } & {c_{,2,,4} } cr
{c_{,3,,1} } & {c_{,3,,2} } & {2c_{,3,,3} } & {c_{,3,,4} } cr
{c_{,4,,1} } & {c_{,4,,2} } & {c_{,4,,3} } & {2c_{,4,,4} } cr
} } right)left( {matrix{ 1 cr 1 cr 1 cr 1 cr
} } right) = left( {matrix{ {n_{;1} } cr {n_{;2} } cr {n_{;3} } cr {n_{;4} } cr
} } right)quad left| {;c_{,j,,k} = c_{,k,,j} } right.
$$
which is a diophantine linear system of four equations in ten unknowns,
which are required to be non-negative integers.
We are interested in the number of solutions of such a system.
It is not an easy task, as you can see from the refered link.
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
Let me introduce what is just a first step.
We shall first of all assume that
$$
n_{;1} + n_{;2} + n_{;3} + n_{;4} = 2m
$$
Then, since order doesn't matter, we are going to arrange the couples alphabetically
$$
left[ {underbrace {aa, ldots ,aa}_{c_{,1,,1} },underbrace {ab, ldots ,ab}_{c_{,1,,2} },
cdots ,underbrace {bb, ldots }_{c_{,2,,2} }, cdots ; cdots ,cd,underbrace {dd, ldots ,dd}_{c_{,4,,4} }} right]
$$
and indicate the number of each different couple by $c_{k,j}$ as shown.
Therefore the symmetric matrix made up by them as shown, shall satisfy
$$
left( {matrix{
{2c_{,1,,1} } & {c_{,1,,2} } & {c_{,1,,3} } & {c_{,1,,4} } cr
{c_{,2,,1} } & {2c_{,2,,2} } & {c_{,2,,3} } & {c_{,2,,4} } cr
{c_{,3,,1} } & {c_{,3,,2} } & {2c_{,3,,3} } & {c_{,3,,4} } cr
{c_{,4,,1} } & {c_{,4,,2} } & {c_{,4,,3} } & {2c_{,4,,4} } cr
} } right)left( {matrix{ 1 cr 1 cr 1 cr 1 cr
} } right) = left( {matrix{ {n_{;1} } cr {n_{;2} } cr {n_{;3} } cr {n_{;4} } cr
} } right)quad left| {;c_{,j,,k} = c_{,k,,j} } right.
$$
which is a diophantine linear system of four equations in ten unknowns,
which are required to be non-negative integers.
We are interested in the number of solutions of such a system.
It is not an easy task, as you can see from the refered link.
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
Let me introduce what is just a first step.
We shall first of all assume that
$$
n_{;1} + n_{;2} + n_{;3} + n_{;4} = 2m
$$
Then, since order doesn't matter, we are going to arrange the couples alphabetically
$$
left[ {underbrace {aa, ldots ,aa}_{c_{,1,,1} },underbrace {ab, ldots ,ab}_{c_{,1,,2} },
cdots ,underbrace {bb, ldots }_{c_{,2,,2} }, cdots ; cdots ,cd,underbrace {dd, ldots ,dd}_{c_{,4,,4} }} right]
$$
and indicate the number of each different couple by $c_{k,j}$ as shown.
Therefore the symmetric matrix made up by them as shown, shall satisfy
$$
left( {matrix{
{2c_{,1,,1} } & {c_{,1,,2} } & {c_{,1,,3} } & {c_{,1,,4} } cr
{c_{,2,,1} } & {2c_{,2,,2} } & {c_{,2,,3} } & {c_{,2,,4} } cr
{c_{,3,,1} } & {c_{,3,,2} } & {2c_{,3,,3} } & {c_{,3,,4} } cr
{c_{,4,,1} } & {c_{,4,,2} } & {c_{,4,,3} } & {2c_{,4,,4} } cr
} } right)left( {matrix{ 1 cr 1 cr 1 cr 1 cr
} } right) = left( {matrix{ {n_{;1} } cr {n_{;2} } cr {n_{;3} } cr {n_{;4} } cr
} } right)quad left| {;c_{,j,,k} = c_{,k,,j} } right.
$$
which is a diophantine linear system of four equations in ten unknowns,
which are required to be non-negative integers.
We are interested in the number of solutions of such a system.
It is not an easy task, as you can see from the refered link.
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
$endgroup$
Let me introduce what is just a first step.
We shall first of all assume that
$$
n_{;1} + n_{;2} + n_{;3} + n_{;4} = 2m
$$
Then, since order doesn't matter, we are going to arrange the couples alphabetically
$$
left[ {underbrace {aa, ldots ,aa}_{c_{,1,,1} },underbrace {ab, ldots ,ab}_{c_{,1,,2} },
cdots ,underbrace {bb, ldots }_{c_{,2,,2} }, cdots ; cdots ,cd,underbrace {dd, ldots ,dd}_{c_{,4,,4} }} right]
$$
and indicate the number of each different couple by $c_{k,j}$ as shown.
Therefore the symmetric matrix made up by them as shown, shall satisfy
$$
left( {matrix{
{2c_{,1,,1} } & {c_{,1,,2} } & {c_{,1,,3} } & {c_{,1,,4} } cr
{c_{,2,,1} } & {2c_{,2,,2} } & {c_{,2,,3} } & {c_{,2,,4} } cr
{c_{,3,,1} } & {c_{,3,,2} } & {2c_{,3,,3} } & {c_{,3,,4} } cr
{c_{,4,,1} } & {c_{,4,,2} } & {c_{,4,,3} } & {2c_{,4,,4} } cr
} } right)left( {matrix{ 1 cr 1 cr 1 cr 1 cr
} } right) = left( {matrix{ {n_{;1} } cr {n_{;2} } cr {n_{;3} } cr {n_{;4} } cr
} } right)quad left| {;c_{,j,,k} = c_{,k,,j} } right.
$$
which is a diophantine linear system of four equations in ten unknowns,
which are required to be non-negative integers.
We are interested in the number of solutions of such a system.
It is not an easy task, as you can see from the refered link.
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
edited Jan 22 at 21:38
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
answered Jan 22 at 21:18
G CabG Cab
19.9k31340
19.9k31340
add a comment |
add a comment |
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$begingroup$
The fact that the items are indistinct makes this question a little vague. You could make the argument that the answer is 10 regardless of what $n_1,ldots,n_4$ are: $aa,bb,cc,dd,ab,ac,ad,bc,bd,cd$. Unless you are asking in how many ways you can pair up ALL of the elements?
$endgroup$
– Nick Peterson
Jan 22 at 19:09
$begingroup$
would like to get your feedback on the answer I provided.
$endgroup$
– G Cab
Jan 28 at 22:26