Mixing
Number of couple of pairs in two datasets
$begingroup$
This is a problem related to combination math.
I have two sets of elements : $A = big(a,b,...,z big)$ and $B = (1,2,...,26)$.
Each element from set $A$ can be paired with an element of set $B$, leading to a pair. I want to make couple of unique pairs so that each pair can be selected multiple times but a couple can only be picked once:
{(a,1), (b,2)} is one valid couple,
{(a,2)(b,2)} is not (2 is used twice in the couple).
{(b,2)(a,1)} is not valid because identical to {(a,1), (b,2)}.
{(a,1),(b,3)} is valid (a,1) already picked but (b,3) not picked.
How many couple of pairs can I make for this dataset?
How many triplets of pairs (per ex.: {(a,2)(b,2),(c,3)}?
If dataset $A$ is of size $m$, and dataset $B$ of size $n$, the formula I get on the paper is $S = sum_{i=0}^{n-1}mBig(2.(n-1)-2i)Big)$, but I experimentally I count $2.binom{n}{2}^2$ combinations.
Event if the second formula is correct, I'm really interested in the general formula (triplets, quadruplets,... of pairs).
combinatorics
asked Jan 31 at 18:30
T4nT413T4nT413
32
$endgroup$
add a comment |
$begingroup$
This is a problem related to combination math.
I have two sets of elements : $A = big(a,b,...,z big)$ and $B = (1,2,...,26)$.
Each element from set $A$ can be paired with an element of set $B$, leading to a pair. I want to make couple of unique pairs so that each pair can be selected multiple times but a couple can only be picked once:
{(a,1), (b,2)} is one valid couple,
{(a,2)(b,2)} is not (2 is used twice in the couple).
{(b,2)(a,1)} is not valid because identical to {(a,1), (b,2)}.
{(a,1),(b,3)} is valid (a,1) already picked but (b,3) not picked.
How many couple of pairs can I make for this dataset?
How many triplets of pairs (per ex.: {(a,2)(b,2),(c,3)}?
If dataset $A$ is of size $m$, and dataset $B$ of size $n$, the formula I get on the paper is $S = sum_{i=0}^{n-1}mBig(2.(n-1)-2i)Big)$, but I experimentally I count $2.binom{n}{2}^2$ combinations.
Event if the second formula is correct, I'm really interested in the general formula (triplets, quadruplets,... of pairs).
combinatorics
asked Jan 31 at 18:30
T4nT413T4nT413
32
$endgroup$
$begingroup$
If I understand your question correctly, each ordered pair must contain a letter and a number in that order and each couple must contain two ordered pairs that do not share the same number or same letter. Is that correct?
$endgroup$
– N. F. Taussig
Feb 1 at 10:25
$begingroup$
@N.F.Taussig Correct!
$endgroup$
– T4nT413
Feb 1 at 10:47
add a comment |
$begingroup$
This is a problem related to combination math.
I have two sets of elements : $A = big(a,b,...,z big)$ and $B = (1,2,...,26)$.
Each element from set $A$ can be paired with an element of set $B$, leading to a pair. I want to make couple of unique pairs so that each pair can be selected multiple times but a couple can only be picked once:
{(a,1), (b,2)} is one valid couple,
{(a,2)(b,2)} is not (2 is used twice in the couple).
{(b,2)(a,1)} is not valid because identical to {(a,1), (b,2)}.
{(a,1),(b,3)} is valid (a,1) already picked but (b,3) not picked.
How many couple of pairs can I make for this dataset?
How many triplets of pairs (per ex.: {(a,2)(b,2),(c,3)}?
If dataset $A$ is of size $m$, and dataset $B$ of size $n$, the formula I get on the paper is $S = sum_{i=0}^{n-1}mBig(2.(n-1)-2i)Big)$, but I experimentally I count $2.binom{n}{2}^2$ combinations.
Event if the second formula is correct, I'm really interested in the general formula (triplets, quadruplets,... of pairs).
combinatorics
asked Jan 31 at 18:30
T4nT413T4nT413
32
$endgroup$
This is a problem related to combination math.
I have two sets of elements : $A = big(a,b,...,z big)$ and $B = (1,2,...,26)$.
Each element from set $A$ can be paired with an element of set $B$, leading to a pair. I want to make couple of unique pairs so that each pair can be selected multiple times but a couple can only be picked once:
{(a,1), (b,2)} is one valid couple,
{(a,2)(b,2)} is not (2 is used twice in the couple).
{(b,2)(a,1)} is not valid because identical to {(a,1), (b,2)}.
{(a,1),(b,3)} is valid (a,1) already picked but (b,3) not picked.
How many couple of pairs can I make for this dataset?
How many triplets of pairs (per ex.: {(a,2)(b,2),(c,3)}?
If dataset $A$ is of size $m$, and dataset $B$ of size $n$, the formula I get on the paper is $S = sum_{i=0}^{n-1}mBig(2.(n-1)-2i)Big)$, but I experimentally I count $2.binom{n}{2}^2$ combinations.
Event if the second formula is correct, I'm really interested in the general formula (triplets, quadruplets,... of pairs).
combinatorics
combinatorics
asked Jan 31 at 18:30
T4nT413T4nT413
32
asked Jan 31 at 18:30
T4nT413T4nT413
32
edited Feb 1 at 17:03
T4nT413
asked Jan 31 at 18:30
T4nT413T4nT413
32
asked Jan 31 at 18:30
T4nT413T4nT413
32
asked Jan 31 at 18:30
T4nT413T4nT413
32
32
$begingroup$
If I understand your question correctly, each ordered pair must contain a letter and a number in that order and each couple must contain two ordered pairs that do not share the same number or same letter. Is that correct?
$endgroup$
– N. F. Taussig
Feb 1 at 10:25
$begingroup$
@N.F.Taussig Correct!
$endgroup$
– T4nT413
Feb 1 at 10:47
add a comment |
$begingroup$
If I understand your question correctly, each ordered pair must contain a letter and a number in that order and each couple must contain two ordered pairs that do not share the same number or same letter. Is that correct?
$endgroup$
– N. F. Taussig
Feb 1 at 10:25
$begingroup$
@N.F.Taussig Correct!
$endgroup$
– T4nT413
Feb 1 at 10:47
$begingroup$
If I understand your question correctly, each ordered pair must contain a letter and a number in that order and each couple must contain two ordered pairs that do not share the same number or same letter. Is that correct?
$endgroup$
– N. F. Taussig
Feb 1 at 10:25
$begingroup$
If I understand your question correctly, each ordered pair must contain a letter and a number in that order and each couple must contain two ordered pairs that do not share the same number or same letter. Is that correct?
$endgroup$
– N. F. Taussig
Feb 1 at 10:25
$begingroup$
@N.F.Taussig Correct!
$endgroup$
– T4nT413
Feb 1 at 10:47
$begingroup$
@N.F.Taussig Correct!
$endgroup$
– T4nT413
Feb 1 at 10:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can make $2cdotbinom{26}2^2$ couples. Here is the reasoning:
Choose two element from A and two elements from B. That gives $binom{26}2^2$. There are $2$ ways to arrange them. e.g. with $(a,b)$ and $(1,2)$, you can do ${(a,1),(b,2)}$ or ${(a,2),(b,1)}$.
answered Feb 1 at 10:29
abc...abc...
3,237739
$endgroup$
1
$begingroup$
So If I understand correctly, for 2 datasets of respective size n and m, we have $2.binom{n}{2}.binom{m}{2}$ ?
$endgroup$
– T4nT413
Feb 1 at 12:21
$begingroup$
@T4nT413 yes that's right.
$endgroup$
– abc...
Feb 2 at 1:02
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can make $2cdotbinom{26}2^2$ couples. Here is the reasoning:
Choose two element from A and two elements from B. That gives $binom{26}2^2$. There are $2$ ways to arrange them. e.g. with $(a,b)$ and $(1,2)$, you can do ${(a,1),(b,2)}$ or ${(a,2),(b,1)}$.
answered Feb 1 at 10:29
abc...abc...
3,237739
$endgroup$
1
$begingroup$
So If I understand correctly, for 2 datasets of respective size n and m, we have $2.binom{n}{2}.binom{m}{2}$ ?
$endgroup$
– T4nT413
Feb 1 at 12:21
$begingroup$
@T4nT413 yes that's right.
$endgroup$
– abc...
Feb 2 at 1:02
add a comment |
$begingroup$
You can make $2cdotbinom{26}2^2$ couples. Here is the reasoning:
Choose two element from A and two elements from B. That gives $binom{26}2^2$. There are $2$ ways to arrange them. e.g. with $(a,b)$ and $(1,2)$, you can do ${(a,1),(b,2)}$ or ${(a,2),(b,1)}$.
answered Feb 1 at 10:29
abc...abc...
3,237739
$endgroup$
1
$begingroup$
So If I understand correctly, for 2 datasets of respective size n and m, we have $2.binom{n}{2}.binom{m}{2}$ ?
$endgroup$
– T4nT413
Feb 1 at 12:21
$begingroup$
@T4nT413 yes that's right.
$endgroup$
– abc...
Feb 2 at 1:02
add a comment |
$begingroup$
You can make $2cdotbinom{26}2^2$ couples. Here is the reasoning:
Choose two element from A and two elements from B. That gives $binom{26}2^2$. There are $2$ ways to arrange them. e.g. with $(a,b)$ and $(1,2)$, you can do ${(a,1),(b,2)}$ or ${(a,2),(b,1)}$.
answered Feb 1 at 10:29
abc...abc...
3,237739
$endgroup$
You can make $2cdotbinom{26}2^2$ couples. Here is the reasoning:
Choose two element from A and two elements from B. That gives $binom{26}2^2$. There are $2$ ways to arrange them. e.g. with $(a,b)$ and $(1,2)$, you can do ${(a,1),(b,2)}$ or ${(a,2),(b,1)}$.
answered Feb 1 at 10:29
abc...abc...
3,237739
answered Feb 1 at 10:29
abc...abc...
3,237739
answered Feb 1 at 10:29
abc...abc...
3,237739
answered Feb 1 at 10:29
abc...abc...
3,237739
3,237739
1
$begingroup$
So If I understand correctly, for 2 datasets of respective size n and m, we have $2.binom{n}{2}.binom{m}{2}$ ?
$endgroup$
– T4nT413
Feb 1 at 12:21
$begingroup$
@T4nT413 yes that's right.
$endgroup$
– abc...
Feb 2 at 1:02
add a comment |
1
$begingroup$
So If I understand correctly, for 2 datasets of respective size n and m, we have $2.binom{n}{2}.binom{m}{2}$ ?
$endgroup$
– T4nT413
Feb 1 at 12:21
$begingroup$
@T4nT413 yes that's right.
$endgroup$
– abc...
Feb 2 at 1:02
1
1
$begingroup$
So If I understand correctly, for 2 datasets of respective size n and m, we have $2.binom{n}{2}.binom{m}{2}$ ?
$endgroup$
– T4nT413
Feb 1 at 12:21
$begingroup$
So If I understand correctly, for 2 datasets of respective size n and m, we have $2.binom{n}{2}.binom{m}{2}$ ?
$endgroup$
– T4nT413
Feb 1 at 12:21
$begingroup$
@T4nT413 yes that's right.
$endgroup$
– abc...
Feb 2 at 1:02
$begingroup$
@T4nT413 yes that's right.
$endgroup$
– abc...
Feb 2 at 1:02
add a comment |
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$begingroup$
If I understand your question correctly, each ordered pair must contain a letter and a number in that order and each couple must contain two ordered pairs that do not share the same number or same letter. Is that correct?
$endgroup$
– N. F. Taussig
Feb 1 at 10:25
$begingroup$
@N.F.Taussig Correct!
$endgroup$
– T4nT413
Feb 1 at 10:47