Mixing
Pairs of non-empty disjoint sets
$begingroup$
Question:
Let $S={1,2, cdots, 10 }$. Then the number of pairs $(A, B)$, where $A$ and
$B$ are non-empty disjoint subsets of $S$ is?
[I could solve the question as demonstrated below, but it involves calculating a tedious sum of products, which would take up a lot of time. It is typically expected to solve this in a couple of minutes, so I was wondering if there was a faster way to do this.]
My approach:
Let the set $A$ consist of $x$ elements. There are ${10 choose x}$ ways of making that selection.
We are now left with $10-x$ elements.
Let the set $B$ consist of $y$ elements. This selection can be done by ${10-x choose y}$ ways.
The total number of ways can be found out by summing over the product of the two above as $sum_{x=1}^{9} sum_{y=1}^{10-x} {10 choose x}{10-x choose y} $ which comes out to be $57002$
combinatorics discrete-mathematics elementary-set-theory
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
$endgroup$
add a comment |
$begingroup$
Question:
Let $S={1,2, cdots, 10 }$. Then the number of pairs $(A, B)$, where $A$ and
$B$ are non-empty disjoint subsets of $S$ is?
[I could solve the question as demonstrated below, but it involves calculating a tedious sum of products, which would take up a lot of time. It is typically expected to solve this in a couple of minutes, so I was wondering if there was a faster way to do this.]
My approach:
Let the set $A$ consist of $x$ elements. There are ${10 choose x}$ ways of making that selection.
We are now left with $10-x$ elements.
Let the set $B$ consist of $y$ elements. This selection can be done by ${10-x choose y}$ ways.
The total number of ways can be found out by summing over the product of the two above as $sum_{x=1}^{9} sum_{y=1}^{10-x} {10 choose x}{10-x choose y} $ which comes out to be $57002$
combinatorics discrete-mathematics elementary-set-theory
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
$endgroup$
2
$begingroup$
Note: this is problem #9 on the 2002 AIME II. (Well, aside from the nonsense that all AIME problems tack onto the end where you need to convert the answer to an integer between $0$ and $999$.)
$endgroup$
– Misha Lavrov
Apr 21 '18 at 5:28
$begingroup$
@MishaLavrov Ah yes, I see! I was not aware of AIME, I'll admit. Thanks
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:31
add a comment |
$begingroup$
Question:
Let $S={1,2, cdots, 10 }$. Then the number of pairs $(A, B)$, where $A$ and
$B$ are non-empty disjoint subsets of $S$ is?
[I could solve the question as demonstrated below, but it involves calculating a tedious sum of products, which would take up a lot of time. It is typically expected to solve this in a couple of minutes, so I was wondering if there was a faster way to do this.]
My approach:
Let the set $A$ consist of $x$ elements. There are ${10 choose x}$ ways of making that selection.
We are now left with $10-x$ elements.
Let the set $B$ consist of $y$ elements. This selection can be done by ${10-x choose y}$ ways.
The total number of ways can be found out by summing over the product of the two above as $sum_{x=1}^{9} sum_{y=1}^{10-x} {10 choose x}{10-x choose y} $ which comes out to be $57002$
combinatorics discrete-mathematics elementary-set-theory
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
$endgroup$
Question:
Let $S={1,2, cdots, 10 }$. Then the number of pairs $(A, B)$, where $A$ and
$B$ are non-empty disjoint subsets of $S$ is?
[I could solve the question as demonstrated below, but it involves calculating a tedious sum of products, which would take up a lot of time. It is typically expected to solve this in a couple of minutes, so I was wondering if there was a faster way to do this.]
My approach:
Let the set $A$ consist of $x$ elements. There are ${10 choose x}$ ways of making that selection.
We are now left with $10-x$ elements.
Let the set $B$ consist of $y$ elements. This selection can be done by ${10-x choose y}$ ways.
The total number of ways can be found out by summing over the product of the two above as $sum_{x=1}^{9} sum_{y=1}^{10-x} {10 choose x}{10-x choose y} $ which comes out to be $57002$
combinatorics discrete-mathematics elementary-set-theory
combinatorics discrete-mathematics elementary-set-theory
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
edited Jan 30 at 17:21
Maria Mazur
49.6k1361124
49.6k1361124
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
asked Apr 21 '18 at 4:54
Tamojit MaitiTamojit Maiti
323313
323313
2
$begingroup$
Note: this is problem #9 on the 2002 AIME II. (Well, aside from the nonsense that all AIME problems tack onto the end where you need to convert the answer to an integer between $0$ and $999$.)
$endgroup$
– Misha Lavrov
Apr 21 '18 at 5:28
$begingroup$
@MishaLavrov Ah yes, I see! I was not aware of AIME, I'll admit. Thanks
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:31
add a comment |
2
$begingroup$
Note: this is problem #9 on the 2002 AIME II. (Well, aside from the nonsense that all AIME problems tack onto the end where you need to convert the answer to an integer between $0$ and $999$.)
$endgroup$
– Misha Lavrov
Apr 21 '18 at 5:28
$begingroup$
@MishaLavrov Ah yes, I see! I was not aware of AIME, I'll admit. Thanks
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:31
2
2
$begingroup$
Note: this is problem #9 on the 2002 AIME II. (Well, aside from the nonsense that all AIME problems tack onto the end where you need to convert the answer to an integer between $0$ and $999$.)
$endgroup$
– Misha Lavrov
Apr 21 '18 at 5:28
$begingroup$
Note: this is problem #9 on the 2002 AIME II. (Well, aside from the nonsense that all AIME problems tack onto the end where you need to convert the answer to an integer between $0$ and $999$.)
$endgroup$
– Misha Lavrov
Apr 21 '18 at 5:28
$begingroup$
@MishaLavrov Ah yes, I see! I was not aware of AIME, I'll admit. Thanks
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:31
$begingroup$
@MishaLavrov Ah yes, I see! I was not aware of AIME, I'll admit. Thanks
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose we have set $A$, $B$ and $S={1,2,...10}$. So for each $xin S$ you can put it in exactly one of this sets: $A$ or in $B$ or $(Acup B)'$. So for each $x$ you have 3 choices and thus you can choose $A,B$ on $3^{10}$ ways.
Now we have to substract all pairs where one of the sets is empty. If $A$ is empty, $B$ could be any subset apart from $A$. So we have $2^{10}$ choices. The same is true if $B$ is empty. So we have $2^{11}$ such pair of sets. But we have pair $(emptyset,emptyset )$ counted twice, we have to substract $2^{11}-1$ bad pairs of sets.
So the finaly result is $3^{10}-2^{11}+1$
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
1
$begingroup$
It will be $3^{10}$, since $A$ has 10 elements
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:14
1
$begingroup$
It cannot be $3^{10}$.You need to ensure that $A,B$ are non empty.
$endgroup$
– copper.hat
Apr 21 '18 at 5:14
2
$begingroup$
$A,B$ can be empty in $2^{11}$ ways. And in one of these instances, both $A,B$ are empty, so we need to add $1$ to the result.
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:16
$begingroup$
The final result becomes $3^{10} - 2^{11} + 1 = 57002$
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:18
1
$begingroup$
Edit...........
$endgroup$
– Maria Mazur
Apr 21 '18 at 5:50
add a comment |
$begingroup$
It is also interesting to note that the sum you get can be simplified in the following way using the binomial theorem:
begin{align}
sum_{x=1}^{9}sum_{y=1}^{10-x}{ 10 choose x}{ 10-x choose y} &= sum_{x=1}^{9}{ 10 choose x}left(sum_{y=0}^{10-x}{ 10-x choose y} -1 right)\
&=sum_{x=1}^{9}{ 10 choose x} (2^{10-x} -1) \
&= sum_{x=0}^{10}{ 10 choose x}2^{10-x} -2^{10}-1 - sum_{x=0}^{10}{ 10 choose x}+1+1 \
&= 3^{10}-2^{10}-1-2^{10}+1+1 \
&= 3^{10}-2^{11}+1
end{align}
Thus getting a pretty interesting identity.
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose we have set $A$, $B$ and $S={1,2,...10}$. So for each $xin S$ you can put it in exactly one of this sets: $A$ or in $B$ or $(Acup B)'$. So for each $x$ you have 3 choices and thus you can choose $A,B$ on $3^{10}$ ways.
Now we have to substract all pairs where one of the sets is empty. If $A$ is empty, $B$ could be any subset apart from $A$. So we have $2^{10}$ choices. The same is true if $B$ is empty. So we have $2^{11}$ such pair of sets. But we have pair $(emptyset,emptyset )$ counted twice, we have to substract $2^{11}-1$ bad pairs of sets.
So the finaly result is $3^{10}-2^{11}+1$
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
1
$begingroup$
It will be $3^{10}$, since $A$ has 10 elements
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:14
1
$begingroup$
It cannot be $3^{10}$.You need to ensure that $A,B$ are non empty.
$endgroup$
– copper.hat
Apr 21 '18 at 5:14
2
$begingroup$
$A,B$ can be empty in $2^{11}$ ways. And in one of these instances, both $A,B$ are empty, so we need to add $1$ to the result.
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:16
$begingroup$
The final result becomes $3^{10} - 2^{11} + 1 = 57002$
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:18
1
$begingroup$
Edit...........
$endgroup$
– Maria Mazur
Apr 21 '18 at 5:50
add a comment |
$begingroup$
Suppose we have set $A$, $B$ and $S={1,2,...10}$. So for each $xin S$ you can put it in exactly one of this sets: $A$ or in $B$ or $(Acup B)'$. So for each $x$ you have 3 choices and thus you can choose $A,B$ on $3^{10}$ ways.
Now we have to substract all pairs where one of the sets is empty. If $A$ is empty, $B$ could be any subset apart from $A$. So we have $2^{10}$ choices. The same is true if $B$ is empty. So we have $2^{11}$ such pair of sets. But we have pair $(emptyset,emptyset )$ counted twice, we have to substract $2^{11}-1$ bad pairs of sets.
So the finaly result is $3^{10}-2^{11}+1$
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
1
$begingroup$
It will be $3^{10}$, since $A$ has 10 elements
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:14
1
$begingroup$
It cannot be $3^{10}$.You need to ensure that $A,B$ are non empty.
$endgroup$
– copper.hat
Apr 21 '18 at 5:14
2
$begingroup$
$A,B$ can be empty in $2^{11}$ ways. And in one of these instances, both $A,B$ are empty, so we need to add $1$ to the result.
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:16
$begingroup$
The final result becomes $3^{10} - 2^{11} + 1 = 57002$
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:18
1
$begingroup$
Edit...........
$endgroup$
– Maria Mazur
Apr 21 '18 at 5:50
add a comment |
$begingroup$
Suppose we have set $A$, $B$ and $S={1,2,...10}$. So for each $xin S$ you can put it in exactly one of this sets: $A$ or in $B$ or $(Acup B)'$. So for each $x$ you have 3 choices and thus you can choose $A,B$ on $3^{10}$ ways.
Now we have to substract all pairs where one of the sets is empty. If $A$ is empty, $B$ could be any subset apart from $A$. So we have $2^{10}$ choices. The same is true if $B$ is empty. So we have $2^{11}$ such pair of sets. But we have pair $(emptyset,emptyset )$ counted twice, we have to substract $2^{11}-1$ bad pairs of sets.
So the finaly result is $3^{10}-2^{11}+1$
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
Suppose we have set $A$, $B$ and $S={1,2,...10}$. So for each $xin S$ you can put it in exactly one of this sets: $A$ or in $B$ or $(Acup B)'$. So for each $x$ you have 3 choices and thus you can choose $A,B$ on $3^{10}$ ways.
Now we have to substract all pairs where one of the sets is empty. If $A$ is empty, $B$ could be any subset apart from $A$. So we have $2^{10}$ choices. The same is true if $B$ is empty. So we have $2^{11}$ such pair of sets. But we have pair $(emptyset,emptyset )$ counted twice, we have to substract $2^{11}-1$ bad pairs of sets.
So the finaly result is $3^{10}-2^{11}+1$
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
edited Apr 21 '18 at 5:54
Tamojit Maiti
323313
323313
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
answered Apr 21 '18 at 5:12
Maria MazurMaria Mazur
49.6k1361124
49.6k1361124
1
$begingroup$
It will be $3^{10}$, since $A$ has 10 elements
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:14
1
$begingroup$
It cannot be $3^{10}$.You need to ensure that $A,B$ are non empty.
$endgroup$
– copper.hat
Apr 21 '18 at 5:14
2
$begingroup$
$A,B$ can be empty in $2^{11}$ ways. And in one of these instances, both $A,B$ are empty, so we need to add $1$ to the result.
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:16
$begingroup$
The final result becomes $3^{10} - 2^{11} + 1 = 57002$
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:18
1
$begingroup$
Edit...........
$endgroup$
– Maria Mazur
Apr 21 '18 at 5:50
add a comment |
1
$begingroup$
It will be $3^{10}$, since $A$ has 10 elements
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:14
1
$begingroup$
It cannot be $3^{10}$.You need to ensure that $A,B$ are non empty.
$endgroup$
– copper.hat
Apr 21 '18 at 5:14
2
$begingroup$
$A,B$ can be empty in $2^{11}$ ways. And in one of these instances, both $A,B$ are empty, so we need to add $1$ to the result.
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:16
$begingroup$
The final result becomes $3^{10} - 2^{11} + 1 = 57002$
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:18
1
$begingroup$
Edit...........
$endgroup$
– Maria Mazur
Apr 21 '18 at 5:50
1
1
$begingroup$
It will be $3^{10}$, since $A$ has 10 elements
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:14
$begingroup$
It will be $3^{10}$, since $A$ has 10 elements
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:14
1
1
$begingroup$
It cannot be $3^{10}$.You need to ensure that $A,B$ are non empty.
$endgroup$
– copper.hat
Apr 21 '18 at 5:14
$begingroup$
It cannot be $3^{10}$.You need to ensure that $A,B$ are non empty.
$endgroup$
– copper.hat
Apr 21 '18 at 5:14
2
2
$begingroup$
$A,B$ can be empty in $2^{11}$ ways. And in one of these instances, both $A,B$ are empty, so we need to add $1$ to the result.
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:16
$begingroup$
$A,B$ can be empty in $2^{11}$ ways. And in one of these instances, both $A,B$ are empty, so we need to add $1$ to the result.
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:16
$begingroup$
The final result becomes $3^{10} - 2^{11} + 1 = 57002$
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:18
$begingroup$
The final result becomes $3^{10} - 2^{11} + 1 = 57002$
$endgroup$
– Tamojit Maiti
Apr 21 '18 at 5:18
1
1
$begingroup$
Edit...........
$endgroup$
– Maria Mazur
Apr 21 '18 at 5:50
$begingroup$
Edit...........
$endgroup$
– Maria Mazur
Apr 21 '18 at 5:50
add a comment |
$begingroup$
It is also interesting to note that the sum you get can be simplified in the following way using the binomial theorem:
begin{align}
sum_{x=1}^{9}sum_{y=1}^{10-x}{ 10 choose x}{ 10-x choose y} &= sum_{x=1}^{9}{ 10 choose x}left(sum_{y=0}^{10-x}{ 10-x choose y} -1 right)\
&=sum_{x=1}^{9}{ 10 choose x} (2^{10-x} -1) \
&= sum_{x=0}^{10}{ 10 choose x}2^{10-x} -2^{10}-1 - sum_{x=0}^{10}{ 10 choose x}+1+1 \
&= 3^{10}-2^{10}-1-2^{10}+1+1 \
&= 3^{10}-2^{11}+1
end{align}
Thus getting a pretty interesting identity.
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
$endgroup$
add a comment |
$begingroup$
It is also interesting to note that the sum you get can be simplified in the following way using the binomial theorem:
begin{align}
sum_{x=1}^{9}sum_{y=1}^{10-x}{ 10 choose x}{ 10-x choose y} &= sum_{x=1}^{9}{ 10 choose x}left(sum_{y=0}^{10-x}{ 10-x choose y} -1 right)\
&=sum_{x=1}^{9}{ 10 choose x} (2^{10-x} -1) \
&= sum_{x=0}^{10}{ 10 choose x}2^{10-x} -2^{10}-1 - sum_{x=0}^{10}{ 10 choose x}+1+1 \
&= 3^{10}-2^{10}-1-2^{10}+1+1 \
&= 3^{10}-2^{11}+1
end{align}
Thus getting a pretty interesting identity.
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
$endgroup$
add a comment |
$begingroup$
It is also interesting to note that the sum you get can be simplified in the following way using the binomial theorem:
begin{align}
sum_{x=1}^{9}sum_{y=1}^{10-x}{ 10 choose x}{ 10-x choose y} &= sum_{x=1}^{9}{ 10 choose x}left(sum_{y=0}^{10-x}{ 10-x choose y} -1 right)\
&=sum_{x=1}^{9}{ 10 choose x} (2^{10-x} -1) \
&= sum_{x=0}^{10}{ 10 choose x}2^{10-x} -2^{10}-1 - sum_{x=0}^{10}{ 10 choose x}+1+1 \
&= 3^{10}-2^{10}-1-2^{10}+1+1 \
&= 3^{10}-2^{11}+1
end{align}
Thus getting a pretty interesting identity.
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
$endgroup$
It is also interesting to note that the sum you get can be simplified in the following way using the binomial theorem:
begin{align}
sum_{x=1}^{9}sum_{y=1}^{10-x}{ 10 choose x}{ 10-x choose y} &= sum_{x=1}^{9}{ 10 choose x}left(sum_{y=0}^{10-x}{ 10-x choose y} -1 right)\
&=sum_{x=1}^{9}{ 10 choose x} (2^{10-x} -1) \
&= sum_{x=0}^{10}{ 10 choose x}2^{10-x} -2^{10}-1 - sum_{x=0}^{10}{ 10 choose x}+1+1 \
&= 3^{10}-2^{10}-1-2^{10}+1+1 \
&= 3^{10}-2^{11}+1
end{align}
Thus getting a pretty interesting identity.
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
answered Apr 21 '18 at 5:24
Antoine GiardAntoine Giard
9427
9427
add a comment |
add a comment |
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Note: this is problem #9 on the 2002 AIME II. (Well, aside from the nonsense that all AIME problems tack onto the end where you need to convert the answer to an integer between $0$ and $999$.)
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– Misha Lavrov
Apr 21 '18 at 5:28
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@MishaLavrov Ah yes, I see! I was not aware of AIME, I'll admit. Thanks
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– Tamojit Maiti
Apr 21 '18 at 5:31