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Reference: question

In the question referred, the answer involves counting number of ways in which 12 identical balls are to be put into 3 identical boxes. This count is $19$ as expressed in @ADG's answer and then probability can be found out.

Is there be any general formula for finding this integer partitions. If I take the same example, say, we need to find out number of ways of 12 identical balls are to be put into 3 identical boxes

After spending a good amount of time, the following are my observations.

  1. If we are looking for a general case, like the number of ways $12$ can be written as sum of positive integers, , we can use Hardy Ramanujan Asymptotic Formula which gives an approximation (in this case we get 77 as the count)

  2. However, if we are looking for a specific case, like the number of ways $12$ can be written as sum of $3$ positive integers, then, there is no specific formula. However, we can derive individual formulas for 1,2,3 etc (reference: Chapter 6 of Andrews and Eriksson,) (in this case we get 19 as the count)

Please share if any formula exist and also share any new findings in this context. Please also tell if my assumptions are right .

Community
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Kiran
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1 Answers1

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I asked myself a similar question when frequenting a combinatorics course. In order not to get confused, I distinguished different partitions with the following criteria:

  • Is $0$ allowed as a summand?
  • Is the number of summands fixed?
  • Do we allow repetitions of the same summand?
  • Does the order in which we list the summands count?

Of course, not all combinations of such criteria make sense, but as you can see there are many different types of integer partitions.

I am not at all an expert, but I can make a couple of examples.

If you want to find the number of ways to write $n$ with $k$ summands, allowing both $0$ and repetitions and considering the order, then you get $f(n)= {n+k-1 \choose k-1}$ (there is a funny and easy proof of this).

A recursive formula for the number of partitions of $n$ with exactly $3$ summands $\ge 1$ and allowing for repetitions but not considering the order, can be found using Polya theory and formal power series: $f(n+1)=f(n)+f(n-1)-f(n-3)-f(n-4)+f(n-5)$ for $n \ge 6$.

You can find the number of ways to write $n$ with summands $\ge 1$, allowing for repetitions but without considering the order, through comparing coefficients in the equality of formal power series $\prod_{k \ge 1} \frac{1}{1-x^k} = \sum_{n \ge 0}f(n)x^n$, or, if you want a fixed amount $s$ of summands, from $\prod_{k \ge 1} \frac{1}{1-x^k y} =\sum_{n,s}f(n,s)x^ny^s.$ I am not aware of an explicit formula in neither of the previous cases.

I hope that I have not made too many mistakes and that this helps you.

57Jimmy
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