Inclusion Exclusion Principle with mixed lower and upper bounds
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1
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I know how to solve most types of these problems, but this one is a bit different.
Problem:
John is getting his friend some balloons for his birthday. He can have 4 types of colors (red, blue, green, yellow) for a total of 25 balloons.
• between 1 and 7 red, • between 2 and 11 blue, • at least 4 green, • at most 6 yellow.
How many combinations of balloons satisfy these conditions?
The green balloons throws me off. I know we can adjust the lower bounds for all colors to 0, and subtract from the total number needed because we know we need at least that initial lower bound. So the conditions become:
0 $le$ red $le$ 6
0 $le$ blue $le$ 9
0 $le$ green
0 $le$ yellow $le$ 6
And the sum is now red + green + blue + yellow = 18
If green had an upper bound, I would solve as follows:
(red $le$ 6 $cap$ blue $le$ 9 $cap$ green $le$ upper bound $cap$ yellow $le$ 6) = $S_{total}$ - (red $ge$ 7 $cup$ blue $ge$ 10 $cup$ green $ge$ upper bound + 1 $cup$ yellow $ge$ 7)
And apply inclusion-exclusion principle to the last portion.. But I don't know the upper bound for green.
Does anyone know how I solve this as is?
combinatorics inclusion-exclusion
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
asked Nov 21 at 5:58
Mr.Mips
64
add a comment |
up vote
1
down vote
favorite
I know how to solve most types of these problems, but this one is a bit different.
Problem:
John is getting his friend some balloons for his birthday. He can have 4 types of colors (red, blue, green, yellow) for a total of 25 balloons.
• between 1 and 7 red, • between 2 and 11 blue, • at least 4 green, • at most 6 yellow.
How many combinations of balloons satisfy these conditions?
The green balloons throws me off. I know we can adjust the lower bounds for all colors to 0, and subtract from the total number needed because we know we need at least that initial lower bound. So the conditions become:
0 $le$ red $le$ 6
0 $le$ blue $le$ 9
0 $le$ green
0 $le$ yellow $le$ 6
And the sum is now red + green + blue + yellow = 18
If green had an upper bound, I would solve as follows:
(red $le$ 6 $cap$ blue $le$ 9 $cap$ green $le$ upper bound $cap$ yellow $le$ 6) = $S_{total}$ - (red $ge$ 7 $cup$ blue $ge$ 10 $cup$ green $ge$ upper bound + 1 $cup$ yellow $ge$ 7)
And apply inclusion-exclusion principle to the last portion.. But I don't know the upper bound for green.
Does anyone know how I solve this as is?
combinatorics inclusion-exclusion
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
asked Nov 21 at 5:58
Mr.Mips
64
1
I think you should try using the multinomial theorem for these questions. I have seen people use it to solve these combinatorics kind of questions.
– Tusky
Nov 21 at 6:06
@Tusky I have to use the method I'm trying to use (using inclusion-exclusion)
– Mr.Mips
Nov 21 at 6:17
I added the tag combinatorics since inclusion-exclusion principle problems fall into that broader category. Adding the combinatorics tag will make your question more likely to be seen since there are more combinatorics questions than inclusion-exclusion principle questions.
– N. F. Taussig
Nov 24 at 12:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know how to solve most types of these problems, but this one is a bit different.
Problem:
John is getting his friend some balloons for his birthday. He can have 4 types of colors (red, blue, green, yellow) for a total of 25 balloons.
• between 1 and 7 red, • between 2 and 11 blue, • at least 4 green, • at most 6 yellow.
How many combinations of balloons satisfy these conditions?
The green balloons throws me off. I know we can adjust the lower bounds for all colors to 0, and subtract from the total number needed because we know we need at least that initial lower bound. So the conditions become:
0 $le$ red $le$ 6
0 $le$ blue $le$ 9
0 $le$ green
0 $le$ yellow $le$ 6
And the sum is now red + green + blue + yellow = 18
If green had an upper bound, I would solve as follows:
(red $le$ 6 $cap$ blue $le$ 9 $cap$ green $le$ upper bound $cap$ yellow $le$ 6) = $S_{total}$ - (red $ge$ 7 $cup$ blue $ge$ 10 $cup$ green $ge$ upper bound + 1 $cup$ yellow $ge$ 7)
And apply inclusion-exclusion principle to the last portion.. But I don't know the upper bound for green.
Does anyone know how I solve this as is?
combinatorics inclusion-exclusion
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
asked Nov 21 at 5:58
Mr.Mips
64
I know how to solve most types of these problems, but this one is a bit different.
Problem:
John is getting his friend some balloons for his birthday. He can have 4 types of colors (red, blue, green, yellow) for a total of 25 balloons.
• between 1 and 7 red, • between 2 and 11 blue, • at least 4 green, • at most 6 yellow.
How many combinations of balloons satisfy these conditions?
The green balloons throws me off. I know we can adjust the lower bounds for all colors to 0, and subtract from the total number needed because we know we need at least that initial lower bound. So the conditions become:
0 $le$ red $le$ 6
0 $le$ blue $le$ 9
0 $le$ green
0 $le$ yellow $le$ 6
And the sum is now red + green + blue + yellow = 18
If green had an upper bound, I would solve as follows:
(red $le$ 6 $cap$ blue $le$ 9 $cap$ green $le$ upper bound $cap$ yellow $le$ 6) = $S_{total}$ - (red $ge$ 7 $cup$ blue $ge$ 10 $cup$ green $ge$ upper bound + 1 $cup$ yellow $ge$ 7)
And apply inclusion-exclusion principle to the last portion.. But I don't know the upper bound for green.
Does anyone know how I solve this as is?
combinatorics inclusion-exclusion
combinatorics inclusion-exclusion
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
asked Nov 21 at 5:58
Mr.Mips
64
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
asked Nov 21 at 5:58
Mr.Mips
64
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
edited Nov 24 at 12:46
N. F. Taussig
42.9k93254
42.9k93254
asked Nov 21 at 5:58
Mr.Mips
64
asked Nov 21 at 5:58
Mr.Mips
64
asked Nov 21 at 5:58
Mr.Mips
64
64
1
I think you should try using the multinomial theorem for these questions. I have seen people use it to solve these combinatorics kind of questions.
– Tusky
Nov 21 at 6:06
@Tusky I have to use the method I'm trying to use (using inclusion-exclusion)
– Mr.Mips
Nov 21 at 6:17
I added the tag combinatorics since inclusion-exclusion principle problems fall into that broader category. Adding the combinatorics tag will make your question more likely to be seen since there are more combinatorics questions than inclusion-exclusion principle questions.
– N. F. Taussig
Nov 24 at 12:49
add a comment |
1
I think you should try using the multinomial theorem for these questions. I have seen people use it to solve these combinatorics kind of questions.
– Tusky
Nov 21 at 6:06
@Tusky I have to use the method I'm trying to use (using inclusion-exclusion)
– Mr.Mips
Nov 21 at 6:17
I added the tag combinatorics since inclusion-exclusion principle problems fall into that broader category. Adding the combinatorics tag will make your question more likely to be seen since there are more combinatorics questions than inclusion-exclusion principle questions.
– N. F. Taussig
Nov 24 at 12:49
1
1
I think you should try using the multinomial theorem for these questions. I have seen people use it to solve these combinatorics kind of questions.
– Tusky
Nov 21 at 6:06
I think you should try using the multinomial theorem for these questions. I have seen people use it to solve these combinatorics kind of questions.
– Tusky
Nov 21 at 6:06
@Tusky I have to use the method I'm trying to use (using inclusion-exclusion)
– Mr.Mips
Nov 21 at 6:17
@Tusky I have to use the method I'm trying to use (using inclusion-exclusion)
– Mr.Mips
Nov 21 at 6:17
I added the tag combinatorics since inclusion-exclusion principle problems fall into that broader category. Adding the combinatorics tag will make your question more likely to be seen since there are more combinatorics questions than inclusion-exclusion principle questions.
– N. F. Taussig
Nov 24 at 12:49
I added the tag combinatorics since inclusion-exclusion principle problems fall into that broader category. Adding the combinatorics tag will make your question more likely to be seen since there are more combinatorics questions than inclusion-exclusion principle questions.
– N. F. Taussig
Nov 24 at 12:49
add a comment |
1 Answer
1
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0
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Assume John has already selected $2$ blue balloons, $4$ green balloons, and $1$ red balloon. Then he must select $18$ more balloons, of which at most $9$ are blue, at most $6$ are red, and at most $6$ are yellow. Since John can only select $18$ more balloons, at most $18$ are green. We do not need to concern ourselves with this restriction since John can use as many green balloons as he likes.
If we let $b$, $g$, $r$, and $y$ denote the number of additional balloons John selects, then we need to find the number of solutions of the equation
$$b + g + r + y = 18 tag{1}$$
in the nonnegative integers subject to the restrictions $b leq 9$, $r leq 6$, $y leq 6$. While we could impose the restriction $g leq 18$ as well, the number of distributions in which $g geq 19$ is zero.
Can you proceed?
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assume John has already selected $2$ blue balloons, $4$ green balloons, and $1$ red balloon. Then he must select $18$ more balloons, of which at most $9$ are blue, at most $6$ are red, and at most $6$ are yellow. Since John can only select $18$ more balloons, at most $18$ are green. We do not need to concern ourselves with this restriction since John can use as many green balloons as he likes.
If we let $b$, $g$, $r$, and $y$ denote the number of additional balloons John selects, then we need to find the number of solutions of the equation
$$b + g + r + y = 18 tag{1}$$
in the nonnegative integers subject to the restrictions $b leq 9$, $r leq 6$, $y leq 6$. While we could impose the restriction $g leq 18$ as well, the number of distributions in which $g geq 19$ is zero.
Can you proceed?
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
add a comment |
up vote
0
down vote
Assume John has already selected $2$ blue balloons, $4$ green balloons, and $1$ red balloon. Then he must select $18$ more balloons, of which at most $9$ are blue, at most $6$ are red, and at most $6$ are yellow. Since John can only select $18$ more balloons, at most $18$ are green. We do not need to concern ourselves with this restriction since John can use as many green balloons as he likes.
If we let $b$, $g$, $r$, and $y$ denote the number of additional balloons John selects, then we need to find the number of solutions of the equation
$$b + g + r + y = 18 tag{1}$$
in the nonnegative integers subject to the restrictions $b leq 9$, $r leq 6$, $y leq 6$. While we could impose the restriction $g leq 18$ as well, the number of distributions in which $g geq 19$ is zero.
Can you proceed?
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
add a comment |
up vote
0
down vote
up vote
0
down vote
Assume John has already selected $2$ blue balloons, $4$ green balloons, and $1$ red balloon. Then he must select $18$ more balloons, of which at most $9$ are blue, at most $6$ are red, and at most $6$ are yellow. Since John can only select $18$ more balloons, at most $18$ are green. We do not need to concern ourselves with this restriction since John can use as many green balloons as he likes.
If we let $b$, $g$, $r$, and $y$ denote the number of additional balloons John selects, then we need to find the number of solutions of the equation
$$b + g + r + y = 18 tag{1}$$
in the nonnegative integers subject to the restrictions $b leq 9$, $r leq 6$, $y leq 6$. While we could impose the restriction $g leq 18$ as well, the number of distributions in which $g geq 19$ is zero.
Can you proceed?
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
Assume John has already selected $2$ blue balloons, $4$ green balloons, and $1$ red balloon. Then he must select $18$ more balloons, of which at most $9$ are blue, at most $6$ are red, and at most $6$ are yellow. Since John can only select $18$ more balloons, at most $18$ are green. We do not need to concern ourselves with this restriction since John can use as many green balloons as he likes.
If we let $b$, $g$, $r$, and $y$ denote the number of additional balloons John selects, then we need to find the number of solutions of the equation
$$b + g + r + y = 18 tag{1}$$
in the nonnegative integers subject to the restrictions $b leq 9$, $r leq 6$, $y leq 6$. While we could impose the restriction $g leq 18$ as well, the number of distributions in which $g geq 19$ is zero.
Can you proceed?
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
answered Nov 24 at 12:45
N. F. Taussig
42.9k93254
42.9k93254
add a comment |
add a comment |
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I think you should try using the multinomial theorem for these questions. I have seen people use it to solve these combinatorics kind of questions.
– Tusky
Nov 21 at 6:06
@Tusky I have to use the method I'm trying to use (using inclusion-exclusion)
– Mr.Mips
Nov 21 at 6:17
I added the tag combinatorics since inclusion-exclusion principle problems fall into that broader category. Adding the combinatorics tag will make your question more likely to be seen since there are more combinatorics questions than inclusion-exclusion principle questions.
– N. F. Taussig
Nov 24 at 12:49