Variation of distributing $k$ balls into $n$ distinguishable boxes
$begingroup$
I am interested in understanding a variation problem of distributing balls into boxes. It seems to be not any of the individual case mentioned in twelvefold way classification.
The problem is described as follows:
In total there are $K$ balls, there are $I$ distinguishable groups by color, and in each color group the balls are indistinguishable with number to be $n_i$ where $i = 1, 2, ldots, I$. Meanwhile, we have $N$ distinguishable boxes where $N$ is always bigger than any $n_i$. Therefore, how many ways are there to distribute these $K$ balls into $N$ boxes, when no box can contain more than one ball from same indistinguishable color group and no empty boxes?
Any suggestion will be appreciated.
combinatorics discrete-mathematics
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
$endgroup$
add a comment |
$begingroup$
I am interested in understanding a variation problem of distributing balls into boxes. It seems to be not any of the individual case mentioned in twelvefold way classification.
The problem is described as follows:
In total there are $K$ balls, there are $I$ distinguishable groups by color, and in each color group the balls are indistinguishable with number to be $n_i$ where $i = 1, 2, ldots, I$. Meanwhile, we have $N$ distinguishable boxes where $N$ is always bigger than any $n_i$. Therefore, how many ways are there to distribute these $K$ balls into $N$ boxes, when no box can contain more than one ball from same indistinguishable color group and no empty boxes?
Any suggestion will be appreciated.
combinatorics discrete-mathematics
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
$endgroup$
add a comment |
$begingroup$
I am interested in understanding a variation problem of distributing balls into boxes. It seems to be not any of the individual case mentioned in twelvefold way classification.
The problem is described as follows:
In total there are $K$ balls, there are $I$ distinguishable groups by color, and in each color group the balls are indistinguishable with number to be $n_i$ where $i = 1, 2, ldots, I$. Meanwhile, we have $N$ distinguishable boxes where $N$ is always bigger than any $n_i$. Therefore, how many ways are there to distribute these $K$ balls into $N$ boxes, when no box can contain more than one ball from same indistinguishable color group and no empty boxes?
Any suggestion will be appreciated.
combinatorics discrete-mathematics
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
$endgroup$
I am interested in understanding a variation problem of distributing balls into boxes. It seems to be not any of the individual case mentioned in twelvefold way classification.
The problem is described as follows:
In total there are $K$ balls, there are $I$ distinguishable groups by color, and in each color group the balls are indistinguishable with number to be $n_i$ where $i = 1, 2, ldots, I$. Meanwhile, we have $N$ distinguishable boxes where $N$ is always bigger than any $n_i$. Therefore, how many ways are there to distribute these $K$ balls into $N$ boxes, when no box can contain more than one ball from same indistinguishable color group and no empty boxes?
Any suggestion will be appreciated.
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
edited Dec 3 '18 at 14:45
Björn Friedrich
2,63961831
2,63961831
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
asked Dec 3 '18 at 11:44
Shaohong BaiShaohong Bai
234
234
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a straightforward application of the principle of inclusion-exclusion. First, count up all the ways to put all the balls into boxes, ignoring the condition that no box can be empty. This is simply
$$
prod_{i=1}^I binom{N}{n_i}
$$
Next, you have to subtract out the "bad" cases where some box is empty. For each of the $N$ boxes, there are $prod_{i=1}^I binom{N-1}{n_i}$ ways to place the balls where that box is empty, and you subtract this for each box, so we are left with
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}
$$
However, distributions with two empty boxes were subtracted out twice by the above formula, so they must be added back in. The result is
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}+binom{N}2prod_{i=1}^I binom{N-2}{n_i}
$$
The summation continues developing in this way. The end result is
$$
boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
$$
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
$endgroup$
$begingroup$
I think this is right answer, I did wrong due to the fact that ignoring the cases it subtracted twice. For other's information, $j$ is the number of empty boxes, where $max(j) = N - max(n_i)$
$endgroup$
– Shaohong Bai
Dec 3 '18 at 16:49
add a comment |
$begingroup$
I got $$P = prod_{i=1}^{I} {Nchoose n_i} - prod_{i=1}^{I} {{N-1}choose n_i}$$
The first part is obviously the combinations of "no box contains more than one ball of the color".
The second part is "no box contains more than one ball of the color AND there is at least one empty box".
This is the Venn diagram I was going by:
The pink part is what we're looking for.
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
$endgroup$
$begingroup$
I think you mean 'The second part is "at least one box contains more than one ball of the color AND there is at least one empty box"'?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:40
$begingroup$
Why would it be "at least one box contains more than one ball of the color"? The second part is distributing one color to $N-1$ boxes, doing it $I$ times.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:41
$begingroup$
Understand what you try to remove from all possible cases. However, I thought the opposite cases to 'no box can contain more than one ball from same indistinguishable color group and no empty boxes' should be 'at least one box contains more than one ball of the color AND there is at least one empty box'. Otherwise, there are might still cases, where a box has multiple same colour balls while there is one or more empty boxes. Maybe you explain a bit more why you think the dropped cases cover all possibilities?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:50
$begingroup$
if you want to use opposite cases, it should be ALL$-$ OPPOSITE CASES. Which is The rectangle $-$ the white part in the Venn diagram.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:55
$begingroup$
I will use $C(A)$ to denote the number of combinations under condition $A$. So what you're trying to find has two conditions, call them $A$(**No box contains more than one ball of the same color**) and $! B$ (NO empty boxes), so the number of combinations is $C(Awedge ! B)$. The pink ellipse is $C(A)$ and the small white ellipse is $C(B)$, to get $C(Awedge !B)$ we can do $C(A) - C(B)$.
$endgroup$
– EvanHehehe
Dec 3 '18 at 15:04
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a straightforward application of the principle of inclusion-exclusion. First, count up all the ways to put all the balls into boxes, ignoring the condition that no box can be empty. This is simply
$$
prod_{i=1}^I binom{N}{n_i}
$$
Next, you have to subtract out the "bad" cases where some box is empty. For each of the $N$ boxes, there are $prod_{i=1}^I binom{N-1}{n_i}$ ways to place the balls where that box is empty, and you subtract this for each box, so we are left with
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}
$$
However, distributions with two empty boxes were subtracted out twice by the above formula, so they must be added back in. The result is
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}+binom{N}2prod_{i=1}^I binom{N-2}{n_i}
$$
The summation continues developing in this way. The end result is
$$
boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
$$
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
$endgroup$
$begingroup$
I think this is right answer, I did wrong due to the fact that ignoring the cases it subtracted twice. For other's information, $j$ is the number of empty boxes, where $max(j) = N - max(n_i)$
$endgroup$
– Shaohong Bai
Dec 3 '18 at 16:49
add a comment |
$begingroup$
This is a straightforward application of the principle of inclusion-exclusion. First, count up all the ways to put all the balls into boxes, ignoring the condition that no box can be empty. This is simply
$$
prod_{i=1}^I binom{N}{n_i}
$$
Next, you have to subtract out the "bad" cases where some box is empty. For each of the $N$ boxes, there are $prod_{i=1}^I binom{N-1}{n_i}$ ways to place the balls where that box is empty, and you subtract this for each box, so we are left with
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}
$$
However, distributions with two empty boxes were subtracted out twice by the above formula, so they must be added back in. The result is
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}+binom{N}2prod_{i=1}^I binom{N-2}{n_i}
$$
The summation continues developing in this way. The end result is
$$
boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
$$
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
$endgroup$
$begingroup$
I think this is right answer, I did wrong due to the fact that ignoring the cases it subtracted twice. For other's information, $j$ is the number of empty boxes, where $max(j) = N - max(n_i)$
$endgroup$
– Shaohong Bai
Dec 3 '18 at 16:49
add a comment |
$begingroup$
This is a straightforward application of the principle of inclusion-exclusion. First, count up all the ways to put all the balls into boxes, ignoring the condition that no box can be empty. This is simply
$$
prod_{i=1}^I binom{N}{n_i}
$$
Next, you have to subtract out the "bad" cases where some box is empty. For each of the $N$ boxes, there are $prod_{i=1}^I binom{N-1}{n_i}$ ways to place the balls where that box is empty, and you subtract this for each box, so we are left with
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}
$$
However, distributions with two empty boxes were subtracted out twice by the above formula, so they must be added back in. The result is
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}+binom{N}2prod_{i=1}^I binom{N-2}{n_i}
$$
The summation continues developing in this way. The end result is
$$
boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
$$
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
$endgroup$
This is a straightforward application of the principle of inclusion-exclusion. First, count up all the ways to put all the balls into boxes, ignoring the condition that no box can be empty. This is simply
$$
prod_{i=1}^I binom{N}{n_i}
$$
Next, you have to subtract out the "bad" cases where some box is empty. For each of the $N$ boxes, there are $prod_{i=1}^I binom{N-1}{n_i}$ ways to place the balls where that box is empty, and you subtract this for each box, so we are left with
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}
$$
However, distributions with two empty boxes were subtracted out twice by the above formula, so they must be added back in. The result is
$$
prod_{i=1}^I binom{N}{n_i}-Ncdot prod_{i=1}^I binom{N-1}{n_i}+binom{N}2prod_{i=1}^I binom{N-2}{n_i}
$$
The summation continues developing in this way. The end result is
$$
boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
$$
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
answered Dec 3 '18 at 16:09
Mike EarnestMike Earnest
21.3k11951
21.3k11951
$begingroup$
I think this is right answer, I did wrong due to the fact that ignoring the cases it subtracted twice. For other's information, $j$ is the number of empty boxes, where $max(j) = N - max(n_i)$
$endgroup$
– Shaohong Bai
Dec 3 '18 at 16:49
add a comment |
$begingroup$
I think this is right answer, I did wrong due to the fact that ignoring the cases it subtracted twice. For other's information, $j$ is the number of empty boxes, where $max(j) = N - max(n_i)$
$endgroup$
– Shaohong Bai
Dec 3 '18 at 16:49
$begingroup$
I think this is right answer, I did wrong due to the fact that ignoring the cases it subtracted twice. For other's information, $j$ is the number of empty boxes, where $max(j) = N - max(n_i)$
$endgroup$
– Shaohong Bai
Dec 3 '18 at 16:49
$begingroup$
I think this is right answer, I did wrong due to the fact that ignoring the cases it subtracted twice. For other's information, $j$ is the number of empty boxes, where $max(j) = N - max(n_i)$
$endgroup$
– Shaohong Bai
Dec 3 '18 at 16:49
add a comment |
$begingroup$
I got $$P = prod_{i=1}^{I} {Nchoose n_i} - prod_{i=1}^{I} {{N-1}choose n_i}$$
The first part is obviously the combinations of "no box contains more than one ball of the color".
The second part is "no box contains more than one ball of the color AND there is at least one empty box".
This is the Venn diagram I was going by:
The pink part is what we're looking for.
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
$endgroup$
$begingroup$
I think you mean 'The second part is "at least one box contains more than one ball of the color AND there is at least one empty box"'?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:40
$begingroup$
Why would it be "at least one box contains more than one ball of the color"? The second part is distributing one color to $N-1$ boxes, doing it $I$ times.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:41
$begingroup$
Understand what you try to remove from all possible cases. However, I thought the opposite cases to 'no box can contain more than one ball from same indistinguishable color group and no empty boxes' should be 'at least one box contains more than one ball of the color AND there is at least one empty box'. Otherwise, there are might still cases, where a box has multiple same colour balls while there is one or more empty boxes. Maybe you explain a bit more why you think the dropped cases cover all possibilities?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:50
$begingroup$
if you want to use opposite cases, it should be ALL$-$ OPPOSITE CASES. Which is The rectangle $-$ the white part in the Venn diagram.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:55
$begingroup$
I will use $C(A)$ to denote the number of combinations under condition $A$. So what you're trying to find has two conditions, call them $A$(**No box contains more than one ball of the same color**) and $! B$ (NO empty boxes), so the number of combinations is $C(Awedge ! B)$. The pink ellipse is $C(A)$ and the small white ellipse is $C(B)$, to get $C(Awedge !B)$ we can do $C(A) - C(B)$.
$endgroup$
– EvanHehehe
Dec 3 '18 at 15:04
|
show 1 more comment
$begingroup$
I got $$P = prod_{i=1}^{I} {Nchoose n_i} - prod_{i=1}^{I} {{N-1}choose n_i}$$
The first part is obviously the combinations of "no box contains more than one ball of the color".
The second part is "no box contains more than one ball of the color AND there is at least one empty box".
This is the Venn diagram I was going by:
The pink part is what we're looking for.
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
$endgroup$
$begingroup$
I think you mean 'The second part is "at least one box contains more than one ball of the color AND there is at least one empty box"'?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:40
$begingroup$
Why would it be "at least one box contains more than one ball of the color"? The second part is distributing one color to $N-1$ boxes, doing it $I$ times.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:41
$begingroup$
Understand what you try to remove from all possible cases. However, I thought the opposite cases to 'no box can contain more than one ball from same indistinguishable color group and no empty boxes' should be 'at least one box contains more than one ball of the color AND there is at least one empty box'. Otherwise, there are might still cases, where a box has multiple same colour balls while there is one or more empty boxes. Maybe you explain a bit more why you think the dropped cases cover all possibilities?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:50
$begingroup$
if you want to use opposite cases, it should be ALL$-$ OPPOSITE CASES. Which is The rectangle $-$ the white part in the Venn diagram.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:55
$begingroup$
I will use $C(A)$ to denote the number of combinations under condition $A$. So what you're trying to find has two conditions, call them $A$(**No box contains more than one ball of the same color**) and $! B$ (NO empty boxes), so the number of combinations is $C(Awedge ! B)$. The pink ellipse is $C(A)$ and the small white ellipse is $C(B)$, to get $C(Awedge !B)$ we can do $C(A) - C(B)$.
$endgroup$
– EvanHehehe
Dec 3 '18 at 15:04
|
show 1 more comment
$begingroup$
I got $$P = prod_{i=1}^{I} {Nchoose n_i} - prod_{i=1}^{I} {{N-1}choose n_i}$$
The first part is obviously the combinations of "no box contains more than one ball of the color".
The second part is "no box contains more than one ball of the color AND there is at least one empty box".
This is the Venn diagram I was going by:
The pink part is what we're looking for.
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
$endgroup$
I got $$P = prod_{i=1}^{I} {Nchoose n_i} - prod_{i=1}^{I} {{N-1}choose n_i}$$
The first part is obviously the combinations of "no box contains more than one ball of the color".
The second part is "no box contains more than one ball of the color AND there is at least one empty box".
This is the Venn diagram I was going by:
The pink part is what we're looking for.
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
edited Dec 3 '18 at 15:07
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
answered Dec 3 '18 at 14:18
EvanHeheheEvanHehehe
1247
1247
$begingroup$
I think you mean 'The second part is "at least one box contains more than one ball of the color AND there is at least one empty box"'?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:40
$begingroup$
Why would it be "at least one box contains more than one ball of the color"? The second part is distributing one color to $N-1$ boxes, doing it $I$ times.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:41
$begingroup$
Understand what you try to remove from all possible cases. However, I thought the opposite cases to 'no box can contain more than one ball from same indistinguishable color group and no empty boxes' should be 'at least one box contains more than one ball of the color AND there is at least one empty box'. Otherwise, there are might still cases, where a box has multiple same colour balls while there is one or more empty boxes. Maybe you explain a bit more why you think the dropped cases cover all possibilities?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:50
$begingroup$
if you want to use opposite cases, it should be ALL$-$ OPPOSITE CASES. Which is The rectangle $-$ the white part in the Venn diagram.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:55
$begingroup$
I will use $C(A)$ to denote the number of combinations under condition $A$. So what you're trying to find has two conditions, call them $A$(**No box contains more than one ball of the same color**) and $! B$ (NO empty boxes), so the number of combinations is $C(Awedge ! B)$. The pink ellipse is $C(A)$ and the small white ellipse is $C(B)$, to get $C(Awedge !B)$ we can do $C(A) - C(B)$.
$endgroup$
– EvanHehehe
Dec 3 '18 at 15:04
|
show 1 more comment
$begingroup$
I think you mean 'The second part is "at least one box contains more than one ball of the color AND there is at least one empty box"'?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:40
$begingroup$
Why would it be "at least one box contains more than one ball of the color"? The second part is distributing one color to $N-1$ boxes, doing it $I$ times.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:41
$begingroup$
Understand what you try to remove from all possible cases. However, I thought the opposite cases to 'no box can contain more than one ball from same indistinguishable color group and no empty boxes' should be 'at least one box contains more than one ball of the color AND there is at least one empty box'. Otherwise, there are might still cases, where a box has multiple same colour balls while there is one or more empty boxes. Maybe you explain a bit more why you think the dropped cases cover all possibilities?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:50
$begingroup$
if you want to use opposite cases, it should be ALL$-$ OPPOSITE CASES. Which is The rectangle $-$ the white part in the Venn diagram.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:55
$begingroup$
I will use $C(A)$ to denote the number of combinations under condition $A$. So what you're trying to find has two conditions, call them $A$(**No box contains more than one ball of the same color**) and $! B$ (NO empty boxes), so the number of combinations is $C(Awedge ! B)$. The pink ellipse is $C(A)$ and the small white ellipse is $C(B)$, to get $C(Awedge !B)$ we can do $C(A) - C(B)$.
$endgroup$
– EvanHehehe
Dec 3 '18 at 15:04
$begingroup$
I think you mean 'The second part is "at least one box contains more than one ball of the color AND there is at least one empty box"'?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:40
$begingroup$
I think you mean 'The second part is "at least one box contains more than one ball of the color AND there is at least one empty box"'?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:40
$begingroup$
Why would it be "at least one box contains more than one ball of the color"? The second part is distributing one color to $N-1$ boxes, doing it $I$ times.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:41
$begingroup$
Why would it be "at least one box contains more than one ball of the color"? The second part is distributing one color to $N-1$ boxes, doing it $I$ times.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:41
$begingroup$
Understand what you try to remove from all possible cases. However, I thought the opposite cases to 'no box can contain more than one ball from same indistinguishable color group and no empty boxes' should be 'at least one box contains more than one ball of the color AND there is at least one empty box'. Otherwise, there are might still cases, where a box has multiple same colour balls while there is one or more empty boxes. Maybe you explain a bit more why you think the dropped cases cover all possibilities?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:50
$begingroup$
Understand what you try to remove from all possible cases. However, I thought the opposite cases to 'no box can contain more than one ball from same indistinguishable color group and no empty boxes' should be 'at least one box contains more than one ball of the color AND there is at least one empty box'. Otherwise, there are might still cases, where a box has multiple same colour balls while there is one or more empty boxes. Maybe you explain a bit more why you think the dropped cases cover all possibilities?
$endgroup$
– Shaohong Bai
Dec 3 '18 at 14:50
$begingroup$
if you want to use opposite cases, it should be ALL$-$ OPPOSITE CASES. Which is The rectangle $-$ the white part in the Venn diagram.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:55
$begingroup$
if you want to use opposite cases, it should be ALL$-$ OPPOSITE CASES. Which is The rectangle $-$ the white part in the Venn diagram.
$endgroup$
– EvanHehehe
Dec 3 '18 at 14:55
$begingroup$
I will use $C(A)$ to denote the number of combinations under condition $A$. So what you're trying to find has two conditions, call them $A$(**No box contains more than one ball of the same color**) and $! B$ (NO empty boxes), so the number of combinations is $C(Awedge ! B)$. The pink ellipse is $C(A)$ and the small white ellipse is $C(B)$, to get $C(Awedge !B)$ we can do $C(A) - C(B)$.
$endgroup$
– EvanHehehe
Dec 3 '18 at 15:04
$begingroup$
I will use $C(A)$ to denote the number of combinations under condition $A$. So what you're trying to find has two conditions, call them $A$(**No box contains more than one ball of the same color**) and $! B$ (NO empty boxes), so the number of combinations is $C(Awedge ! B)$. The pink ellipse is $C(A)$ and the small white ellipse is $C(B)$, to get $C(Awedge !B)$ we can do $C(A) - C(B)$.
$endgroup$
– EvanHehehe
Dec 3 '18 at 15:04
|
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