Question

Three balls are placed at random in three boxes, with no restriction on the number of balls per box. List the 27 possible outcomes, explaining your notation.

--I thought there would be 9 combinations associated with Box 1, 2 and 3 so 27 possibilities... but I could only find 8 possibilities?

Answer 1

I said if Xi = Box 1,2,3 and Bi = Balls 1,2,3
I thought box 1 could have these possibilities:
X1, X1B1, X1B2, X1B3, X1B1B2, X1B1B3, X1B2B3, X1B1B2B3 = 8 combos then multiply by 3 (for X2 and X3), but the problem states there are 27 combos =\

Answer 2

Oh wait I think I have to backwards actually ._.

Answer 3

I. There can be a single ball in each box OR
II. There can be 2 balls in a single box, one ball in another box, one empty OR
III. There can be 3 balls in a single box, 2 boxes empty

right?

Answer 4

Are these all the possibilities?

Answer 5

?

Answer 6

How many ways can I occur? Ans. 1

Answer 7

How many ways can II occur? Ans. 3

Answer 8

Is that right for II?

Answer 9

should be 6. right. 2 balls can be in any of 3 boxes, one ball can be in any of the remaining 2, one in the last. 3*2*1 = 6

Answer 10

What about III?

Answer 11

3 ways?

Answer 12

Total number I get is 1 + 6 + 3 = 10, where order doesn't matter. Things change if it does.

Answer 13

i'm still a bit confused ;s

Answer 14

The thing though is that you cna have empty boxes

Answer 15

If order matters then we have this:
for I - Single ball for the 1st box can be any of the 3 balls, the second box can contain any of the remaining 2 balls, the last 1 way. so 3*2*1 are the number of arrangements, 6.

Answer 16

for II - There can be 3 ways to group 2 balls if order matters. Then the set of 2 balls can be in any of the 3 boxes, for a total of 3*3 = 9.

Answer 17

Also, the 1 ball left can be in any of the remaining 2 boxes, so the total for II is 2*9=18

Answer 18

For III, the 3 balls can only be grouped one way and can be in any of the 3 boxes, so there are 3 ways.

Answer 19

So, the identity of the balls matter, there are 6+18+3= 27 arrangements possible.

Do you see this?

Answer 20

Think of the balls as numbered, then their identities will matter.

Answer 21

This is more challenging since you need to keep track of two things that can permute. The balls themselves can be grouped and the boxes that they are in is another permutation.

Answer 22

If you separate and treat each by them selves you will get an answer. These then are multiplied because the number of ways is multiplied. Each of one contributes to the other, which is by definition, multiplication.

Answer 23

Let's look at the balls first.

Answer 24

Do you want to continue?