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How many ways are there to split 4 alike red, 5 alike blue and 7 alike black balls among:

a) Two boxes, without any restriction?

b) Two boxes, with no box empty?

c) Three boxes, without any restriction?

d) Three boxes, with no box empty?

I'm relatively unskilled with combinatorics so I would really love it if there could be some reasoning/logic explained as to how these types of problems are solved (for instance, if the numbers 4,5,7 were slightly altered).

N. F. Taussig
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Wanyu Tang
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  • Please indicate what you have tried and where you are stuck. Are you familiar with combinations with repetition? – N. F. Taussig Jan 16 '17 at 23:36
  • @N. F. Taussig What I've tried so far was: assign each colour a letter (red - A, blue - B and black - C). I thought of it as permuting words with AAAABBBBBCCCCCCC, which is 16!/(4!•5!•6!). I can then put in 'dividors' to split into 2 or 3 groups. The solution I get with this method is very large and I'm not sure what is wrong. – Wanyu Tang Jan 17 '17 at 00:47
  • In the first case, there are five ways to distribute the red balls (with $0$, $1$, $2$, $3$, or $4$ balls being placed in the first box, and the rest being placed in the second box), six ways to distribute the blue balls, and eight ways to distribute the black balls, so there are $5 \cdot 6 \cdot 8 = 240$ ways of distributing the balls to two boxes without restriction. These problems are combinations with repetition (see my first comment) rather than permutations of a multiset (the approach you took). – N. F. Taussig Jan 17 '17 at 02:02
  • Are there any good free resources which can explain these types of combinatorics? If so, please recommend some to me. I feel like Wikipedia is a bit unclear/complex for me to understand at a high school student level. – Wanyu Tang Jan 17 '17 at 06:49
  • You may find it useful to read the responses to this question and this one. If I have time later this week, I will address the questions you posed here. – N. F. Taussig Jan 17 '17 at 22:41

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