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I solved the problem using Python language and dynamic programming algorithm, What is the mathematical solution to this problem?

# Given n balls, k buckets, and m is a limit
n = 8
k = 6
m = 2

The number of ways to place j balls into the first i buckets,


where the number of balls placed in each bucket, denoted by 'a', satisfies the condition 1 <= a <= m


dp = [ [0 for i in range(n+1)] for j in range(k+1)]


The bucket numbering starts from 1. Here, we are initializing the first bucket


for j in range(1,m+1):
    dp[1][j] = 1


for i in range(2,k+1):
    for j in range(i,min(m*i,n)+1):
        for z in range(1,m+1):
            if j-z>=i-1:
                dp[i][j] += dp[i-1][j-z]
            else:
                break


print(dp[k][n])
user1315461
  • 39
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 26 '24 at 07:02
  • Check out https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) – Samar Apr 26 '24 at 07:09
  • Thank you very much, I'll go check it right away – user1315461 Apr 26 '24 at 07:14
  • Check this one as well because you need to combine both the ideas to get the answer. https://math.stackexchange.com/questions/89417/number-of-ways-to-put-n-unlabeled-balls-in-k-bins-with-a-max-of-m-balls-in – Samar Apr 26 '24 at 07:36

1 Answers1

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Taking help from this post, Stars and Bars method and extending the question to having at least p balls and at most m balls

this is what you need $$\sum_{i=0}^r (-1)^i {k\choose i} {{n-k(p-1)-1-i(m-p+1)}\choose{k-1}}$$

where $r = {\left \lfloor {\frac{n-kp}{m-p+1}} \right \rfloor}$

Samar
  • 556