How many ways three men split up twelve sodas?

Question

assume it is possible for two to get zero and one to get all

Answer 1

From combinatorics, (n+k-1 choose k) is the formula for the number of ways you can form groups of k out of n objects, allowing each object to be part of a group more than once or not at all (n choose k is defined as n!/(k!*(n-k)!), where n! is the factorial of n, i.e., the product of integers between 1 and n).

In this case, you can reformulate your question into "In how many ways can you form groups of 12 out of three people, allowing each person to be in a group more than once or not at all?". I know it sounds strange for a person to be part of a group more than once, but think about it: a group of 9 times the 1st person, 2 times the 2nd person and 1 time the 3rd person corresponds to splitting up the sodas as 9-2-1 among the 3 people.

Knowing this, you only need to substitute into the magic formula to get (3+12-1 choose 12) = (14 choose 12) = 14!/(12!*2!), however, you can easily verify that 14!/12!=13*14, so the answer is 13*14/(1*2)=13*7=91.

Answer 2

The number is practically infinite, and is limited only by the granularity of the total number of molecules in the sodas.

Answer 3

each gets 4