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?

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 #2!# to account for the identical white marbles and add 1 for being able to allow both white marbles in the same box. The general formula for a combination is:

#C_(n,k)=(n!)/((k)!(n-k)!)# with #n="population", k="picks"#

and so we have:

#C_(4,2)/(2!)+1=(4!)/((2!)(4-2)!(2!))=(4!)/((2!)(2!)(2!))+1=>#

#24/8+1=3+1=4#

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 #2!# to account for the identical white marbles, and adding 1 to account for the two white marbles being in the same box, and then multiplying by 2 for the two boxes:

#2(C_(4,3)/(2!)+1)=2((4!)/((3!)(4-3)!(2!))+1)=>#

#2((4!)/((3!)(1!)(2!))+1)=>#

#2((24/12)+1)=2(2+1)=2(3)=6#

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:

#4+6+2=12#