Grouping Two Different Coloured Objects

Having three black objects B and one white object W they can be grouped in 7 ways like this:
In how many ways can sixty black objects B and forty white objects W be thus grouped?

This problem can be solved by introducing the concept of partitioning in number theory. The partition of a number is a way of writing it as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition.

We have 60 black objects and 40 white objects, and we can look at the problem as: “in how many ways can we partition a number 60 with sums of integers from 1 to 40 (representing the group in which black balls are with some number of white balls and the number of white balls not exceeding 40), plus ways to partition the number 40 with sums of integers from 1 to 60 (representing the group of white balls with some number of black balls and the number of black balls not exceeding 60)”. This will avoid double counting the identical groups.

For partitioning, we use partition numbers P(n, k), which represents the number of partitions of n using integers up to k. We need to add up all the partition numbers from P(60,1) to P(60,40) and all partition numbers from P(40,1) to P(40,60), minus P(40,40) (to take out the case where black and white balls are not grouped together). Partition numbers can be calculated with various methods, including generating functions, recursion relation or explicit formulas.

However, the calculation of partition numbers for such large n and k as in this problem is not trivial and practically not feasible to be done by hand, but with the help of computer algorithms or statistical softwares, it could be efficiently calculated.

Hence, the count of all configurations is P(60,1)+P(60,2)+…+P(60,40) + P(40,1)+…+P(40,60) – P(40,40).

More Answers:
Step Numbers
Consecutive Positive Divisors
Golden Triplets

Error 403 The request cannot be completed because you have exceeded your quota. : quotaExceeded


Recent Posts

Don't Miss Out! Sign Up Now!

Sign up now to get started for free!