Choosing from identical objects of different types

[yucheng]

TL;DR

I am asking about the general formulation. But to be concrete....

How many ways can we choose 6 objects from say {A,A,B,B,B,C,D,E,E,E,F,G,G,G}? (identical objects of different type)

Pure evil: What's the probability of choosing 2A's and 2 B's?

This can be formulated as partitions with constraints or choosing with finite replacement or choosing from identical objects of different types.

Do you have any comments? What's the most general way to solve it? Inclusion-exclusion (PIE) is impractical for general case. Where to read further on this problem?

Here's what I've found.

Hypergeometric distribution???
https://math.stackexchange.com/ques...-out-of-n-identical-objects?noredirect=1&lq=1

Generating functions: this is promising but incomplete. Is brute force expansion the only way to get the coefficient?
https://math.stackexchange.com/ques...replacement-from-a-set-that-contains-duplicat
https://math.stackexchange.com/a/2757736/767174

Generating functions: Perhaps more complete and concrete. But in the end, the author is unable to compute the coefficients (at least by hand) but still suggests PIE
https://math.stackexchange.com/questions/41724/combination-problem-with-constraints

Again PIE (with stars and bars)
https://math.stackexchange.com/questions/3047584/drawing-balls-with-a-finite-number-of-replacement

Slightly more comprehensive, but the author suggests PIE, which kills the brain for slightly more complicated problems
https://math.stackexchange.com/ques...mula-for-combinations-with-identical-elements

I think this is plain wrong!
https://math.stackexchange.com/questions/582788/distinct-combinations-of-non-distinct-elements?rq=1

 

[andrewkirk]

To calculate probabilities I think you need the multivariate hypergeometric distribution, described here: https://en.wikipedia.org/wiki/Hypergeometric_distribution#Multivariate_hypergeometric_distribution
It is a generalisation of the hypergeometric distribution to cases where there are more than two categories (usually described as colours).

To count the "number of ways" we'd first need to define exactly what we mean by that, as the term becomes ambiguous when we have identical items.