Variation of distributing $k$ balls into $n$ distinguishable boxes












3












$begingroup$


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







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 14:45









      Björn Friedrich

      2,63961831




      2,63961831










      asked Dec 3 '18 at 11:44









      Shaohong BaiShaohong Bai

      234




      234






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          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
          $$
          boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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



















          0












          $begingroup$

          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:enter image description here
          The pink part is what we're looking for.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          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
          $$
          boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          2












          $begingroup$

          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
          $$
          boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          2












          2








          2





          $begingroup$

          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
          $$
          boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
          $$






          share|cite|improve this answer









          $endgroup$



          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
          $$
          boxed{sum_{j=0}^N (-1)^jbinom{N}jprod_{i=1}^I binom{N-j}{n_i}.}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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











          0












          $begingroup$

          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:enter image description here
          The pink part is what we're looking for.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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


















          0












          $begingroup$

          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:enter image description here
          The pink part is what we're looking for.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          0












          0








          0





          $begingroup$

          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:enter image description here
          The pink part is what we're looking for.






          share|cite|improve this answer











          $endgroup$



          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:enter image description here
          The pink part is what we're looking for.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 15:07

























          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




















          • $begingroup$
            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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