Permutations arrangement question

Question

Question : three balls are to be placed in 3 different boxes not necessarily with one ball in each box. Any box can hold one two or 3 balls. Find the number of ways 3 balls which are of the same colour and therefore indistinguishable can be arranged.

My solution : I looked at the possibilities for this

3 balls in each box = 3 possibilities

1 ball in a box = 1 possibility

Then 2 balls in 1 box , 1 in another box and one box without any = 6 possibilities

Hence total ways is 10?

I would like to know if I'm correct and if there is an easier way to do this

Answer 1

OK let's try this.

How many balls in the first box? If it is three we are done. There is one way.

If it is two then there is one ball left over. There are two places to put that. Two more ways.

If it is one then we have two balls left over. There are three ways of placing those (1,1),(2,0),(0,2)

If it is none then there are four ways (0,3),(1,2),(2,1) or (3,). This is four more.

In total we have $1+2+3+4$ which I'm going to call $10$.

Hey, we have the same answer so I think you are right! :)

Answer 2

You are correct.

It comes to finding the number of sums $a+b+c=3$ where $a,b,c$ are nonnegative integers.

This can be solved with stars and bars, and gives the outcome $$\binom{3+2}2=10$$

Answer 3

We can solve it using beggars method as well. Take two partitions

O. P O. P O

As the position of partitions change the no. of balls in each segment change and can be distributed among the boxes. Say in this position each box is gonna get one ball each.

O O PP O

In this position 1st box is getting 2, 2nd is getting 0 and 3rd is getting one.

So the answer basically is the no. of ways to arrange the three identical balla and 2 partitions which is C(5,2) =10

Answer 4

I assume an urn must not contain a ball. Then you can use the following formula to calculate the number of arrangements

$$\binom{n+k-1}{k}$$

where

$n$=number of distinguishable urns.

$k$=number of indistinguishable balls

In your case we have

$$\binom{3+3-1}{3}=\binom{5}{3}=\binom{5}{2}=\frac{5\cdot 4}{1\cdot 2}=10$$

It confirms the result of your counting.