# There are 2 identical boxes, 2 identical white balls, 1 red ball, and 1 blue ball. In how many different ways can we fill each box with exactly 2 balls? In how many different ways can we fill one box with 3 balls and the other with 1?

##
In how many different ways can we fill one box with 4 balls and the other with none? In how many different ways can the boxes be filled with any number of balls?

In how many different ways can we fill one box with 4 balls and the other with none? In how many different ways can the boxes be filled with any number of balls?

##### 1 Answer

2 balls to a box = 4 ways. 3 balls in one box = 6 ways. 4 balls in one box = 2 ways. The number of ways to put the marbles into the two boxes = 12.

#### Explanation:

We have 2 boxes and 4 balls (2 white, 1 red, 1 blue) and we have a series of questions about the number of ways we can distribute the balls across the boxes.

**2 balls per box**

In Box A, we can put in 2 marbles:

WW, WR, WB, BR

and the other box will take the other 2 marbles. And so there are 4 ways to do this.

We can do this algebraically by saying that we have a combination with a population of 4, choosing 2, and dividing by

and so we have:

**3 balls in 1 box, 1 in the other**

In Box A, we can put in 3 marbles:

WWR, WWB, WRB

and the other box will take the other marble. And so there are 3 ways of doing this for Box A. There are then also 3 ways to do it for Box B, for a total of 6 ways.

We can do this algebraically by saying we have a combination with a population of 4, choosing 3, dividing by

**4 balls in one box**

There is only 1 way to place all 4 marbles into Box A and 1 way to do it with Box B, so there are 2 ways in total.

**Any number of balls into the boxes - all marbles need to be in boxes**

If we allow for any number of marbles into the boxes and all the marbles must be in boxes, we can simply add the different answers together: