If you have N balls and m containers, how many ways to split the balls into each container?

Question

Assume each container can hold as many balls as you wish, ie All N balls can go into 1 container. Also all the balls must be in a container, ie you cannot have a situation where a ball is in no container.

E.g. If we have 2 balls and 2 containers, then the number of ways to split the balls is 3

2 - 0
0 - 2
1 - 1

Please provide simple explanation for how you derive the answer.

Answer 1

by your example it seems as if the N balls are indistinguishable.

The number of ways to distribute N indistinguishable balls into m distinguishable cells is:

C(m + N - 1, N)

where the C(n,r) notation means: n!/(r!(n - r)!)

so in your example this is:

C(2 + 2 -1,2) = C(3,2) = 3!/(2!1!) = 3

Answer 2

If you assume that balls are different from each other like having different colors:
Every single ball has N ways that it can be placed. Every ball is independent from each other. So you simply multiply number of ways for a ball (N) number of balls (m) times to find all the ways. That is N^m (N to the power m).
For 2 balls and 2 containers assume that the balls are blue and red: 2^2 = 4
red & blue - no balls
red - blue
blue - red
no balls - red & blue
====================================
If the balls are similar then it is more complicated

Answer 3

Hey brother.It is a good question.your question can be solved by using MULTINOMIAL THEOREM.
you may find multinomial theorem explanation on the net.
MULTINOMIAL Theorem can solve your question easily ,it uses the concept of camparing coefficients & powers of variable on both sides .You may find its proof and application in books of higher algebra.(i.e Hall & knight etc.)