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In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty?

What if the boxes can be empty?

N. F. Taussig
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Dominic Joseph
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  • This can probably answer your question: https://math.stackexchange.com/questions/691823/in-how-many-ways-can-20-identical-balls-be-distributed-into-4-distinct-boxes-sub – Charlie Shuffler Apr 11 '19 at 11:25
  • No. Thats identical items in distinct bucket. Different from my question – Dominic Joseph Apr 11 '19 at 11:28
  • Besides here I have 2 seperate groups of items – Dominic Joseph Apr 11 '19 at 11:29
  • You could still use it. Anyway, combine the first link with this one and you should be able to find the answer https://math.stackexchange.com/questions/1659517/in-how-many-ways-can-4-red-balls-and-7-blue-balls-be-arranged-in-3-boxes?rq=1 If you check the questions related to those, there's very many very similar to yours – Charlie Shuffler Apr 11 '19 at 11:32
  • I don't think. I'm well aware of that approach. This one has a different approach. – Dominic Joseph Apr 11 '19 at 11:34
  • Things areeasy when atleast the box or the ball is distinct. When both are identical , the problem gets very difficult. – Dominic Joseph Apr 11 '19 at 11:36
  • I assume you meant to write something along the lines of "In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty?" – N. F. Taussig Apr 11 '19 at 15:18
  • Yes..thats is correct – Dominic Joseph Apr 12 '19 at 03:03
  • Related well-known solutions: (1) $M$ identical balls into $N$ identical boxes is the integer partition problem. (2) Excluding empty boxes can be handled by counting the complement. However, I don't see right away how to handle the fact there are $2$ groups of identical balls. It doesn't seem to be the product of two integer partition problems... – antkam Apr 12 '19 at 13:31

1 Answers1

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Using the Polya Enumeration Theorem we get for the two cases using the cycle index of the symmetric group with empty boxes

$$Q_1 = [R^3 B^4] Z\left(S_3; (1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right)$$

and without

$$Q_2 = [R^3 B^4] Z\left(S_3; -1 + (1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right).$$

Now the cycle index is

$$Z(S_3) = 1/6\,{a_{{1}}}^{3}+1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}.$$

Doing the substitution we find

$$\bbox[5px,border:2px solid #00A000]{ Q_1 = 28 \quad\text{and}\quad Q_2 = 18.}$$

If we want to do these by hand, here is an example. We use the alternate form

$$[R^3 B^4] Z\left(S_3; \frac{1}{1-R}\frac{1}{1-B}\right).$$

We get from the first term of the cycle index

$$[R^3 B^4] \frac{1}{6} \frac{1}{(1-R)^3}\frac{1}{(1-B)^3} = \frac{1}{6} {3+2\choose 2} {4+2\choose 2} = 25.$$

We get from the second term

$$[R^3 B^4] \frac{1}{2} \frac{1}{1-R^2}\frac{1}{1-B^2} \frac{1}{1-R}\frac{1}{1-B} = \frac{1}{2} (1+1)\times (1+1+1) = 3.$$

Here we have e.g. for the coefficient on $B^4$ the possibilities $(B^2)^2 (B^1)^0,$ $(B^2)^1 (B^1)^2$ and $(B^2)^0 (B^1)^4.$

At last we get from the third term

$$[R^3 B^4] \frac{1}{3} \frac{1}{1-R^3}\frac{1}{1-B^3} = 0.$$

Add these to obtain $25+3=28.$

Marko Riedel
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