Mixing
Counting ordered pairs $(X,Y)$
$begingroup$
Consider two disjoint sets $A$ and $B$ with $|A|=|B|=8$. Find two ways to count the number of ordered pairs $(X,Y)$ with $X subseteq A$ and $Y subseteq (B cup X)$ with $|Y|=8$.
Is my guess correct?
- We have $|B cup X| = |B| + |X| - |B cap X| = 8+|X|$ (since $A$ and $B$ are disjoint, $X subseteq A$).
We know that $1 le^* |X| le |A| = 8$ and $|Y| = 8 le 8 + |X|$.
(*We'll leave out $|X| = 0$, since it does not contribute to the number of ordered pairs $(X,Y)$).
This results in $ 16 ge |B cup X| ge 9 $. So for all possible elements for $Y$ we do the following: $sum_{i=9}^{16} binom{i}{8}$. There are from $1$ through $8$ possible elements for $X$ (chosen from $A$): $sum_{k=1}^8 binom{8}{k}$.
The number of ordered pairs is $sum_{k=1}^8 binom{8}{k} cdot sum_{i=9}^{16} binom{i}{8}$.
I'm struggling to find another way.
combinatorics elementary-set-theory
asked Jan 28 at 19:19
ZacharyZachary
1939
$endgroup$
add a comment |
$begingroup$
Consider two disjoint sets $A$ and $B$ with $|A|=|B|=8$. Find two ways to count the number of ordered pairs $(X,Y)$ with $X subseteq A$ and $Y subseteq (B cup X)$ with $|Y|=8$.
Is my guess correct?
- We have $|B cup X| = |B| + |X| - |B cap X| = 8+|X|$ (since $A$ and $B$ are disjoint, $X subseteq A$).
We know that $1 le^* |X| le |A| = 8$ and $|Y| = 8 le 8 + |X|$.
(*We'll leave out $|X| = 0$, since it does not contribute to the number of ordered pairs $(X,Y)$).
This results in $ 16 ge |B cup X| ge 9 $. So for all possible elements for $Y$ we do the following: $sum_{i=9}^{16} binom{i}{8}$. There are from $1$ through $8$ possible elements for $X$ (chosen from $A$): $sum_{k=1}^8 binom{8}{k}$.
The number of ordered pairs is $sum_{k=1}^8 binom{8}{k} cdot sum_{i=9}^{16} binom{i}{8}$.
I'm struggling to find another way.
combinatorics elementary-set-theory
asked Jan 28 at 19:19
ZacharyZachary
1939
$endgroup$
$begingroup$
"couple" is not the right word for (a,b) in english ; it has to be replaced by "ordered pair".
$endgroup$
– Jean Marie
Jan 28 at 19:43
$begingroup$
Thanks! I edited my post. Do you think my answer is correct, @JeanMarie?
$endgroup$
– Zachary
Jan 28 at 19:49
3
$begingroup$
If I compute well, the number of such ordered pairs should be $sum_{k=color{red}{0}}^8 binom{8}{k} cdot binom{8+k}{8}$ (the remark in red : the void set is not forbidden): without a double summation.
$endgroup$
– Jean Marie
Jan 28 at 20:06
add a comment |
$begingroup$
Consider two disjoint sets $A$ and $B$ with $|A|=|B|=8$. Find two ways to count the number of ordered pairs $(X,Y)$ with $X subseteq A$ and $Y subseteq (B cup X)$ with $|Y|=8$.
Is my guess correct?
- We have $|B cup X| = |B| + |X| - |B cap X| = 8+|X|$ (since $A$ and $B$ are disjoint, $X subseteq A$).
We know that $1 le^* |X| le |A| = 8$ and $|Y| = 8 le 8 + |X|$.
(*We'll leave out $|X| = 0$, since it does not contribute to the number of ordered pairs $(X,Y)$).
This results in $ 16 ge |B cup X| ge 9 $. So for all possible elements for $Y$ we do the following: $sum_{i=9}^{16} binom{i}{8}$. There are from $1$ through $8$ possible elements for $X$ (chosen from $A$): $sum_{k=1}^8 binom{8}{k}$.
The number of ordered pairs is $sum_{k=1}^8 binom{8}{k} cdot sum_{i=9}^{16} binom{i}{8}$.
I'm struggling to find another way.
combinatorics elementary-set-theory
asked Jan 28 at 19:19
ZacharyZachary
1939
$endgroup$
Consider two disjoint sets $A$ and $B$ with $|A|=|B|=8$. Find two ways to count the number of ordered pairs $(X,Y)$ with $X subseteq A$ and $Y subseteq (B cup X)$ with $|Y|=8$.
Is my guess correct?
- We have $|B cup X| = |B| + |X| - |B cap X| = 8+|X|$ (since $A$ and $B$ are disjoint, $X subseteq A$).
We know that $1 le^* |X| le |A| = 8$ and $|Y| = 8 le 8 + |X|$.
(*We'll leave out $|X| = 0$, since it does not contribute to the number of ordered pairs $(X,Y)$).
This results in $ 16 ge |B cup X| ge 9 $. So for all possible elements for $Y$ we do the following: $sum_{i=9}^{16} binom{i}{8}$. There are from $1$ through $8$ possible elements for $X$ (chosen from $A$): $sum_{k=1}^8 binom{8}{k}$.
The number of ordered pairs is $sum_{k=1}^8 binom{8}{k} cdot sum_{i=9}^{16} binom{i}{8}$.
I'm struggling to find another way.
combinatorics elementary-set-theory
combinatorics elementary-set-theory
asked Jan 28 at 19:19
ZacharyZachary
1939
asked Jan 28 at 19:19
ZacharyZachary
1939
edited Jan 28 at 19:49
Zachary
asked Jan 28 at 19:19
ZacharyZachary
1939
asked Jan 28 at 19:19
ZacharyZachary
1939
asked Jan 28 at 19:19
ZacharyZachary
1939
1939
$begingroup$
"couple" is not the right word for (a,b) in english ; it has to be replaced by "ordered pair".
$endgroup$
– Jean Marie
Jan 28 at 19:43
$begingroup$
Thanks! I edited my post. Do you think my answer is correct, @JeanMarie?
$endgroup$
– Zachary
Jan 28 at 19:49
3
$begingroup$
If I compute well, the number of such ordered pairs should be $sum_{k=color{red}{0}}^8 binom{8}{k} cdot binom{8+k}{8}$ (the remark in red : the void set is not forbidden): without a double summation.
$endgroup$
– Jean Marie
Jan 28 at 20:06
add a comment |
$begingroup$
"couple" is not the right word for (a,b) in english ; it has to be replaced by "ordered pair".
$endgroup$
– Jean Marie
Jan 28 at 19:43
$begingroup$
Thanks! I edited my post. Do you think my answer is correct, @JeanMarie?
$endgroup$
– Zachary
Jan 28 at 19:49
3
$begingroup$
If I compute well, the number of such ordered pairs should be $sum_{k=color{red}{0}}^8 binom{8}{k} cdot binom{8+k}{8}$ (the remark in red : the void set is not forbidden): without a double summation.
$endgroup$
– Jean Marie
Jan 28 at 20:06
$begingroup$
"couple" is not the right word for (a,b) in english ; it has to be replaced by "ordered pair".
$endgroup$
– Jean Marie
Jan 28 at 19:43
$begingroup$
"couple" is not the right word for (a,b) in english ; it has to be replaced by "ordered pair".
$endgroup$
– Jean Marie
Jan 28 at 19:43
$begingroup$
Thanks! I edited my post. Do you think my answer is correct, @JeanMarie?
$endgroup$
– Zachary
Jan 28 at 19:49
$begingroup$
Thanks! I edited my post. Do you think my answer is correct, @JeanMarie?
$endgroup$
– Zachary
Jan 28 at 19:49
3
3
$begingroup$
If I compute well, the number of such ordered pairs should be $sum_{k=color{red}{0}}^8 binom{8}{k} cdot binom{8+k}{8}$ (the remark in red : the void set is not forbidden): without a double summation.
$endgroup$
– Jean Marie
Jan 28 at 20:06
$begingroup$
If I compute well, the number of such ordered pairs should be $sum_{k=color{red}{0}}^8 binom{8}{k} cdot binom{8+k}{8}$ (the remark in red : the void set is not forbidden): without a double summation.
$endgroup$
– Jean Marie
Jan 28 at 20:06
add a comment |
1 Answer
1
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$begingroup$
Formula: #(A×B) = #A × #B
For each a in A there are #B ways to make a into an ordered pair.
Thus add #B, #A times.
The other way.
For each b in B, there are #A ways to make b ...
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Formula: #(A×B) = #A × #B
For each a in A there are #B ways to make a into an ordered pair.
Thus add #B, #A times.
The other way.
For each b in B, there are #A ways to make b ...
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
$endgroup$
add a comment |
$begingroup$
Formula: #(A×B) = #A × #B
For each a in A there are #B ways to make a into an ordered pair.
Thus add #B, #A times.
The other way.
For each b in B, there are #A ways to make b ...
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
$endgroup$
add a comment |
$begingroup$
Formula: #(A×B) = #A × #B
For each a in A there are #B ways to make a into an ordered pair.
Thus add #B, #A times.
The other way.
For each b in B, there are #A ways to make b ...
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
$endgroup$
Formula: #(A×B) = #A × #B
For each a in A there are #B ways to make a into an ordered pair.
Thus add #B, #A times.
The other way.
For each b in B, there are #A ways to make b ...
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
answered Jan 29 at 0:03
William ElliotWilliam Elliot
8,8882820
8,8882820
add a comment |
add a comment |
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$begingroup$
"couple" is not the right word for (a,b) in english ; it has to be replaced by "ordered pair".
$endgroup$
– Jean Marie
Jan 28 at 19:43
$begingroup$
Thanks! I edited my post. Do you think my answer is correct, @JeanMarie?
$endgroup$
– Zachary
Jan 28 at 19:49
3
$begingroup$
If I compute well, the number of such ordered pairs should be $sum_{k=color{red}{0}}^8 binom{8}{k} cdot binom{8+k}{8}$ (the remark in red : the void set is not forbidden): without a double summation.
$endgroup$
– Jean Marie
Jan 28 at 20:06