How many ways can you distribute 4 different balls among 5 identical boxes?

Guest Aug 23, 2020

#1**0 **

Assuming order matters:

There are 5 cases, {4}, {3,1}, {2, 2}, {2, 1, 1}, and {1, 1, 1, 1}.

{4} - 4! which is 24

{3, 1} - 4 x 3! - 24 (4 ways to pick the single one and 3! ways to rearrange the others)

{2, 2} - 6 x 2 = 12 (6 ways to choose the pairs, and two ways to order the pairs)

{2, 1, 1} - 12 (6 ways to choose the pair and two ways to order it)

{1, 1, 1, 1} - 1 (just one way to order it)

24+24+12+12+1 = 73.

I hope this helped.

Firebolt Aug 23, 2020

#2**0 **

By the way, I found a helpful link - this might be helpful. It uses the same method I used but it explains it better than I did.

https://math.stackexchange.com/questions/1873138/distribution-of-4-distinct-balls-into-5-identical-boxes-without-any-restrictions#:~:text=There%20are%202%20ways%20to,6%C3%972%3D12%20cases.

Firebolt
Aug 23, 2020